35:I[651451,["/_next/static/chunks/2c7cbuj0myjl0.js?dpl=nogit-1789466213","/_next/static/chunks/2juje0t3dlhbm.js?dpl=nogit-1789466213","/_next/static/chunks/19ijf68pshdcn.js?dpl=nogit-1789466213","/_next/static/chunks/04u0lp5-90-_i.js?dpl=nogit-1789466213","/_next/static/chunks/0i-8vweyyd2z1.js?dpl=nogit-1789466213","/_next/static/chunks/0br-rafekpauh.js?dpl=nogit-1789466213","/_next/static/chunks/0u008gt6051u7.js?dpl=nogit-1789466213","/_next/static/chunks/0zkcxwtbsqliz.js?dpl=nogit-1789466213","/_next/static/chunks/1wq0tmwxm0o03.js?dpl=nogit-1789466213","/_next/static/chunks/1bbv7o2-m-0ni.js?dpl=nogit-1789466213","/_next/static/chunks/3n6wyk-4-ou-2.js?dpl=nogit-1789466213","/_next/static/chunks/1h4wccj42vygd.js?dpl=nogit-1789466213","/_next/static/chunks/3jpl-42mbad5n.js?dpl=nogit-1789466213","/_next/static/chunks/1w_cokuayn2xb.js?dpl=nogit-1789466213","/_next/static/chunks/442oofikvq4ao.js?dpl=nogit-1789466213","/_next/static/chunks/2os7nxouq91eq.js?dpl=nogit-1789466213","/_next/static/chunks/3hxsq15e9l6ex.js?dpl=nogit-1789466213","/_next/static/chunks/3ht89vb0p09bm.js?dpl=nogit-1789466213","/_next/static/chunks/2qlkd_50jnb67.js?dpl=nogit-1789466213","/_next/static/chunks/3ml3vooxhiupb.js?dpl=nogit-1789466213","/_next/static/chunks/0ry5u_b6y0dbk.js?dpl=nogit-1789466213","/_next/static/chunks/0qfm0wv_qgrsj.js?dpl=nogit-1789466213","/_next/static/chunks/0gtr6g0fi5l0-.js?dpl=nogit-1789466213","/_next/static/chunks/14jjqpa8a5nws.js?dpl=nogit-1789466213","/_next/static/chunks/0i6x1920m4zk4.js?dpl=nogit-1789466213","/_next/static/chunks/1yts3efoejfo0.js?dpl=nogit-1789466213","/_next/static/chunks/04zr-wqafvwoa.js?dpl=nogit-1789466213","/_next/static/chunks/2id2jbelkqob6.js?dpl=nogit-1789466213","/_next/static/chunks/18kibg6v3ribz.js?dpl=nogit-1789466213","/_next/static/chunks/0k-y_4mcfub-p.js?dpl=nogit-1789466213","/_next/static/chunks/2_zyy4shki--m.js?dpl=nogit-1789466213","/_next/static/chunks/2ydqdr10ndcei.js?dpl=nogit-1789466213","/_next/static/chunks/0fia0hf7-7vfo.js?dpl=nogit-1789466213","/_next/static/chunks/43dq9llw-iyp8.js?dpl=nogit-1789466213","/_next/static/chunks/2dx-vzwyw9a3g.js?dpl=nogit-1789466213","/_next/static/chunks/0a_nnu2-xlcam.js?dpl=nogit-1789466213","/_next/static/chunks/12zn735ys1sga.js?dpl=nogit-1789466213","/_next/static/chunks/0llqhg6-bxn_b.js?dpl=nogit-1789466213","/_next/static/chunks/20xy3ur1zq5mk.js?dpl=nogit-1789466213","/_next/static/chunks/0fkz1t-1q4vvx.js?dpl=nogit-1789466213","/_next/static/chunks/2cu8u4002xlhk.js?dpl=nogit-1789466213","/_next/static/chunks/23dzzhwlmv97g.js?dpl=nogit-1789466213","/_next/static/chunks/2muh31f8m33t4.js?dpl=nogit-1789466213","/_next/static/chunks/0epkvyva8shyj.js?dpl=nogit-1789466213","/_next/static/chunks/03ql4_ajt89dj.js?dpl=nogit-1789466213","/_next/static/chunks/2b6zx8pzxydki.js?dpl=nogit-1789466213","/_next/static/chunks/0tapm_eeixiw6.js?dpl=nogit-1789466213","/_next/static/chunks/1tw7dwog40-j_.js?dpl=nogit-1789466213","/_next/static/chunks/3q_n1v1z_hn1-.js?dpl=nogit-1789466213","/_next/static/chunks/14ynz93rnm6_r.js?dpl=nogit-1789466213","/_next/static/chunks/3zzkpug-bge13.js?dpl=nogit-1789466213","/_next/static/chunks/1ewppg55gz8ao.js?dpl=nogit-1789466213","/_next/static/chunks/2sghpash7yifv.js?dpl=nogit-1789466213","/_next/static/chunks/1r2o7os-xpck9.js?dpl=nogit-1789466213","/_next/static/chunks/3xskuweujaw05.js?dpl=nogit-1789466213","/_next/static/chunks/2y341-r8ncizo.js?dpl=nogit-1789466213","/_next/static/chunks/30dopbgd1dkf9.js?dpl=nogit-1789466213","/_next/static/chunks/3tn81b-gro1f1.js?dpl=nogit-1789466213","/_next/static/chunks/139ut7awapdtx.js?dpl=nogit-1789466213","/_next/static/chunks/0fx374hbtcqun.js?dpl=nogit-1789466213","/_next/static/chunks/3v9ybc6cyn2el.js?dpl=nogit-1789466213","/_next/static/chunks/1ksyn9m68446d.js?dpl=nogit-1789466213","/_next/static/chunks/28z7oiyzc2ir9.js?dpl=nogit-1789466213","/_next/static/chunks/0zg1om_k0m5zv.js?dpl=nogit-1789466213","/_next/static/chunks/1fn8bf7933b7y.js?dpl=nogit-1789466213","/_next/static/chunks/43ajf4ofaeu5y.js?dpl=nogit-1789466213","/_next/static/chunks/258b_3vm29gcy.js?dpl=nogit-1789466213","/_next/static/chunks/2dr--mw_fu802.js?dpl=nogit-1789466213","/_next/static/chunks/14k7sugwbq3i8.js?dpl=nogit-1789466213","/_next/static/chunks/0k562u9txct8-.js?dpl=nogit-1789466213"],"UserProgressProvider"] 36:I[196793,["/_next/static/chunks/2c7cbuj0myjl0.js?dpl=nogit-1789466213","/_next/static/chunks/2juje0t3dlhbm.js?dpl=nogit-1789466213","/_next/static/chunks/19ijf68pshdcn.js?dpl=nogit-1789466213","/_next/static/chunks/04u0lp5-90-_i.js?dpl=nogit-1789466213","/_next/static/chunks/0i-8vweyyd2z1.js?dpl=nogit-1789466213","/_next/static/chunks/0br-rafekpauh.js?dpl=nogit-1789466213","/_next/static/chunks/0u008gt6051u7.js?dpl=nogit-1789466213","/_next/static/chunks/0zkcxwtbsqliz.js?dpl=nogit-1789466213","/_next/static/chunks/1wq0tmwxm0o03.js?dpl=nogit-1789466213","/_next/static/chunks/1bbv7o2-m-0ni.js?dpl=nogit-1789466213","/_next/static/chunks/3n6wyk-4-ou-2.js?dpl=nogit-1789466213","/_next/static/chunks/1h4wccj42vygd.js?dpl=nogit-1789466213","/_next/static/chunks/3jpl-42mbad5n.js?dpl=nogit-1789466213","/_next/static/chunks/1w_cokuayn2xb.js?dpl=nogit-1789466213","/_next/static/chunks/442oofikvq4ao.js?dpl=nogit-1789466213","/_next/static/chunks/2os7nxouq91eq.js?dpl=nogit-1789466213","/_next/static/chunks/3hxsq15e9l6ex.js?dpl=nogit-1789466213","/_next/static/chunks/3ht89vb0p09bm.js?dpl=nogit-1789466213","/_next/static/chunks/2qlkd_50jnb67.js?dpl=nogit-1789466213","/_next/static/chunks/3ml3vooxhiupb.js?dpl=nogit-1789466213","/_next/static/chunks/0ry5u_b6y0dbk.js?dpl=nogit-1789466213","/_next/static/chunks/0qfm0wv_qgrsj.js?dpl=nogit-1789466213","/_next/static/chunks/0gtr6g0fi5l0-.js?dpl=nogit-1789466213","/_next/static/chunks/14jjqpa8a5nws.js?dpl=nogit-1789466213","/_next/static/chunks/0i6x1920m4zk4.js?dpl=nogit-1789466213","/_next/static/chunks/1yts3efoejfo0.js?dpl=nogit-1789466213","/_next/static/chunks/04zr-wqafvwoa.js?dpl=nogit-1789466213","/_next/static/chunks/2id2jbelkqob6.js?dpl=nogit-1789466213","/_next/static/chunks/18kibg6v3ribz.js?dpl=nogit-1789466213","/_next/static/chunks/0k-y_4mcfub-p.js?dpl=nogit-1789466213","/_next/static/chunks/2_zyy4shki--m.js?dpl=nogit-1789466213","/_next/static/chunks/2ydqdr10ndcei.js?dpl=nogit-1789466213","/_next/static/chunks/0fia0hf7-7vfo.js?dpl=nogit-1789466213","/_next/static/chunks/43dq9llw-iyp8.js?dpl=nogit-1789466213","/_next/static/chunks/2dx-vzwyw9a3g.js?dpl=nogit-1789466213","/_next/static/chunks/0a_nnu2-xlcam.js?dpl=nogit-1789466213","/_next/static/chunks/12zn735ys1sga.js?dpl=nogit-1789466213","/_next/static/chunks/0llqhg6-bxn_b.js?dpl=nogit-1789466213","/_next/static/chunks/20xy3ur1zq5mk.js?dpl=nogit-1789466213","/_next/static/chunks/0fkz1t-1q4vvx.js?dpl=nogit-1789466213","/_next/static/chunks/2cu8u4002xlhk.js?dpl=nogit-1789466213","/_next/static/chunks/23dzzhwlmv97g.js?dpl=nogit-1789466213","/_next/static/chunks/2muh31f8m33t4.js?dpl=nogit-1789466213","/_next/static/chunks/0epkvyva8shyj.js?dpl=nogit-1789466213","/_next/static/chunks/03ql4_ajt89dj.js?dpl=nogit-1789466213","/_next/static/chunks/2b6zx8pzxydki.js?dpl=nogit-1789466213","/_next/static/chunks/0tapm_eeixiw6.js?dpl=nogit-1789466213","/_next/static/chunks/1tw7dwog40-j_.js?dpl=nogit-1789466213","/_next/static/chunks/3q_n1v1z_hn1-.js?dpl=nogit-1789466213","/_next/static/chunks/14ynz93rnm6_r.js?dpl=nogit-1789466213","/_next/static/chunks/3zzkpug-bge13.js?dpl=nogit-1789466213","/_next/static/chunks/1ewppg55gz8ao.js?dpl=nogit-1789466213","/_next/static/chunks/2sghpash7yifv.js?dpl=nogit-1789466213","/_next/static/chunks/1r2o7os-xpck9.js?dpl=nogit-1789466213","/_next/static/chunks/3xskuweujaw05.js?dpl=nogit-1789466213","/_next/static/chunks/2y341-r8ncizo.js?dpl=nogit-1789466213","/_next/static/chunks/30dopbgd1dkf9.js?dpl=nogit-1789466213","/_next/static/chunks/3tn81b-gro1f1.js?dpl=nogit-1789466213","/_next/static/chunks/139ut7awapdtx.js?dpl=nogit-1789466213","/_next/static/chunks/0fx374hbtcqun.js?dpl=nogit-1789466213","/_next/static/chunks/3v9ybc6cyn2el.js?dpl=nogit-1789466213","/_next/static/chunks/1ksyn9m68446d.js?dpl=nogit-1789466213","/_next/static/chunks/28z7oiyzc2ir9.js?dpl=nogit-1789466213","/_next/static/chunks/0zg1om_k0m5zv.js?dpl=nogit-1789466213","/_next/static/chunks/1fn8bf7933b7y.js?dpl=nogit-1789466213","/_next/static/chunks/43ajf4ofaeu5y.js?dpl=nogit-1789466213","/_next/static/chunks/258b_3vm29gcy.js?dpl=nogit-1789466213","/_next/static/chunks/2dr--mw_fu802.js?dpl=nogit-1789466213","/_next/static/chunks/14k7sugwbq3i8.js?dpl=nogit-1789466213","/_next/static/chunks/0k562u9txct8-.js?dpl=nogit-1789466213"],"Gingerbread"] 38:I[716297,["/_next/static/chunks/2c7cbuj0myjl0.js?dpl=nogit-1789466213","/_next/static/chunks/2juje0t3dlhbm.js?dpl=nogit-1789466213","/_next/static/chunks/19ijf68pshdcn.js?dpl=nogit-1789466213","/_next/static/chunks/04u0lp5-90-_i.js?dpl=nogit-1789466213","/_next/static/chunks/0i-8vweyyd2z1.js?dpl=nogit-1789466213","/_next/static/chunks/0br-rafekpauh.js?dpl=nogit-1789466213","/_next/static/chunks/0u008gt6051u7.js?dpl=nogit-1789466213","/_next/static/chunks/0zkcxwtbsqliz.js?dpl=nogit-1789466213","/_next/static/chunks/1wq0tmwxm0o03.js?dpl=nogit-1789466213","/_next/static/chunks/1bbv7o2-m-0ni.js?dpl=nogit-1789466213","/_next/static/chunks/3n6wyk-4-ou-2.js?dpl=nogit-1789466213","/_next/static/chunks/1h4wccj42vygd.js?dpl=nogit-1789466213","/_next/static/chunks/3jpl-42mbad5n.js?dpl=nogit-1789466213","/_next/static/chunks/1w_cokuayn2xb.js?dpl=nogit-1789466213","/_next/static/chunks/442oofikvq4ao.js?dpl=nogit-1789466213","/_next/static/chunks/2os7nxouq91eq.js?dpl=nogit-1789466213","/_next/static/chunks/3hxsq15e9l6ex.js?dpl=nogit-1789466213","/_next/static/chunks/3ht89vb0p09bm.js?dpl=nogit-1789466213","/_next/static/chunks/2qlkd_50jnb67.js?dpl=nogit-1789466213","/_next/static/chunks/3ml3vooxhiupb.js?dpl=nogit-1789466213","/_next/static/chunks/0ry5u_b6y0dbk.js?dpl=nogit-1789466213","/_next/static/chunks/0qfm0wv_qgrsj.js?dpl=nogit-1789466213","/_next/static/chunks/0gtr6g0fi5l0-.js?dpl=nogit-1789466213","/_next/static/chunks/14jjqpa8a5nws.js?dpl=nogit-1789466213","/_next/static/chunks/0i6x1920m4zk4.js?dpl=nogit-1789466213","/_next/static/chunks/1yts3efoejfo0.js?dpl=nogit-1789466213","/_next/static/chunks/04zr-wqafvwoa.js?dpl=nogit-1789466213","/_next/static/chunks/2id2jbelkqob6.js?dpl=nogit-1789466213","/_next/static/chunks/18kibg6v3ribz.js?dpl=nogit-1789466213","/_next/static/chunks/0k-y_4mcfub-p.js?dpl=nogit-1789466213","/_next/static/chunks/2_zyy4shki--m.js?dpl=nogit-1789466213","/_next/static/chunks/2ydqdr10ndcei.js?dpl=nogit-1789466213","/_next/static/chunks/0fia0hf7-7vfo.js?dpl=nogit-1789466213","/_next/static/chunks/43dq9llw-iyp8.js?dpl=nogit-1789466213","/_next/static/chunks/2dx-vzwyw9a3g.js?dpl=nogit-1789466213","/_next/static/chunks/0a_nnu2-xlcam.js?dpl=nogit-1789466213","/_next/static/chunks/12zn735ys1sga.js?dpl=nogit-1789466213","/_next/static/chunks/0llqhg6-bxn_b.js?dpl=nogit-1789466213","/_next/static/chunks/20xy3ur1zq5mk.js?dpl=nogit-1789466213","/_next/static/chunks/0fkz1t-1q4vvx.js?dpl=nogit-1789466213","/_next/static/chunks/2cu8u4002xlhk.js?dpl=nogit-1789466213","/_next/static/chunks/23dzzhwlmv97g.js?dpl=nogit-1789466213","/_next/static/chunks/2muh31f8m33t4.js?dpl=nogit-1789466213","/_next/static/chunks/0epkvyva8shyj.js?dpl=nogit-1789466213","/_next/static/chunks/03ql4_ajt89dj.js?dpl=nogit-1789466213","/_next/static/chunks/2b6zx8pzxydki.js?dpl=nogit-1789466213","/_next/static/chunks/0tapm_eeixiw6.js?dpl=nogit-1789466213","/_next/static/chunks/1tw7dwog40-j_.js?dpl=nogit-1789466213","/_next/static/chunks/3q_n1v1z_hn1-.js?dpl=nogit-1789466213","/_next/static/chunks/14ynz93rnm6_r.js?dpl=nogit-1789466213","/_next/static/chunks/3zzkpug-bge13.js?dpl=nogit-1789466213","/_next/static/chunks/1ewppg55gz8ao.js?dpl=nogit-1789466213","/_next/static/chunks/2sghpash7yifv.js?dpl=nogit-1789466213","/_next/static/chunks/1r2o7os-xpck9.js?dpl=nogit-1789466213","/_next/static/chunks/3xskuweujaw05.js?dpl=nogit-1789466213","/_next/static/chunks/2y341-r8ncizo.js?dpl=nogit-1789466213","/_next/static/chunks/30dopbgd1dkf9.js?dpl=nogit-1789466213","/_next/static/chunks/3tn81b-gro1f1.js?dpl=nogit-1789466213","/_next/static/chunks/139ut7awapdtx.js?dpl=nogit-1789466213","/_next/static/chunks/0fx374hbtcqun.js?dpl=nogit-1789466213","/_next/static/chunks/3v9ybc6cyn2el.js?dpl=nogit-1789466213","/_next/static/chunks/1ksyn9m68446d.js?dpl=nogit-1789466213","/_next/static/chunks/28z7oiyzc2ir9.js?dpl=nogit-1789466213","/_next/static/chunks/0zg1om_k0m5zv.js?dpl=nogit-1789466213","/_next/static/chunks/1fn8bf7933b7y.js?dpl=nogit-1789466213","/_next/static/chunks/43ajf4ofaeu5y.js?dpl=nogit-1789466213","/_next/static/chunks/258b_3vm29gcy.js?dpl=nogit-1789466213","/_next/static/chunks/2dr--mw_fu802.js?dpl=nogit-1789466213","/_next/static/chunks/14k7sugwbq3i8.js?dpl=nogit-1789466213","/_next/static/chunks/0k562u9txct8-.js?dpl=nogit-1789466213"],"TopicRecentlyViewed"] 39:I[647935,["/_next/static/chunks/2c7cbuj0myjl0.js?dpl=nogit-1789466213","/_next/static/chunks/2juje0t3dlhbm.js?dpl=nogit-1789466213","/_next/static/chunks/19ijf68pshdcn.js?dpl=nogit-1789466213","/_next/static/chunks/04u0lp5-90-_i.js?dpl=nogit-1789466213","/_next/static/chunks/0i-8vweyyd2z1.js?dpl=nogit-1789466213","/_next/static/chunks/0br-rafekpauh.js?dpl=nogit-1789466213","/_next/static/chunks/0u008gt6051u7.js?dpl=nogit-1789466213","/_next/static/chunks/0zkcxwtbsqliz.js?dpl=nogit-1789466213","/_next/static/chunks/1wq0tmwxm0o03.js?dpl=nogit-1789466213","/_next/static/chunks/1bbv7o2-m-0ni.js?dpl=nogit-1789466213","/_next/static/chunks/3n6wyk-4-ou-2.js?dpl=nogit-1789466213","/_next/static/chunks/1h4wccj42vygd.js?dpl=nogit-1789466213","/_next/static/chunks/3jpl-42mbad5n.js?dpl=nogit-1789466213","/_next/static/chunks/1w_cokuayn2xb.js?dpl=nogit-1789466213","/_next/static/chunks/442oofikvq4ao.js?dpl=nogit-1789466213","/_next/static/chunks/2os7nxouq91eq.js?dpl=nogit-1789466213","/_next/static/chunks/3hxsq15e9l6ex.js?dpl=nogit-1789466213","/_next/static/chunks/3ht89vb0p09bm.js?dpl=nogit-1789466213","/_next/static/chunks/2qlkd_50jnb67.js?dpl=nogit-1789466213","/_next/static/chunks/3ml3vooxhiupb.js?dpl=nogit-1789466213","/_next/static/chunks/0ry5u_b6y0dbk.js?dpl=nogit-1789466213","/_next/static/chunks/0qfm0wv_qgrsj.js?dpl=nogit-1789466213","/_next/static/chunks/0gtr6g0fi5l0-.js?dpl=nogit-1789466213","/_next/static/chunks/14jjqpa8a5nws.js?dpl=nogit-1789466213","/_next/static/chunks/0i6x1920m4zk4.js?dpl=nogit-1789466213","/_next/static/chunks/1yts3efoejfo0.js?dpl=nogit-1789466213","/_next/static/chunks/04zr-wqafvwoa.js?dpl=nogit-1789466213","/_next/static/chunks/2id2jbelkqob6.js?dpl=nogit-1789466213","/_next/static/chunks/18kibg6v3ribz.js?dpl=nogit-1789466213","/_next/static/chunks/0k-y_4mcfub-p.js?dpl=nogit-1789466213","/_next/static/chunks/2_zyy4shki--m.js?dpl=nogit-1789466213","/_next/static/chunks/2ydqdr10ndcei.js?dpl=nogit-1789466213","/_next/static/chunks/0fia0hf7-7vfo.js?dpl=nogit-1789466213","/_next/static/chunks/43dq9llw-iyp8.js?dpl=nogit-1789466213","/_next/static/chunks/2dx-vzwyw9a3g.js?dpl=nogit-1789466213","/_next/static/chunks/0a_nnu2-xlcam.js?dpl=nogit-1789466213","/_next/static/chunks/12zn735ys1sga.js?dpl=nogit-1789466213","/_next/static/chunks/0llqhg6-bxn_b.js?dpl=nogit-1789466213","/_next/static/chunks/20xy3ur1zq5mk.js?dpl=nogit-1789466213","/_next/static/chunks/0fkz1t-1q4vvx.js?dpl=nogit-1789466213","/_next/static/chunks/2cu8u4002xlhk.js?dpl=nogit-1789466213","/_next/static/chunks/23dzzhwlmv97g.js?dpl=nogit-1789466213","/_next/static/chunks/2muh31f8m33t4.js?dpl=nogit-1789466213","/_next/static/chunks/0epkvyva8shyj.js?dpl=nogit-1789466213","/_next/static/chunks/03ql4_ajt89dj.js?dpl=nogit-1789466213","/_next/static/chunks/2b6zx8pzxydki.js?dpl=nogit-1789466213","/_next/static/chunks/0tapm_eeixiw6.js?dpl=nogit-1789466213","/_next/static/chunks/1tw7dwog40-j_.js?dpl=nogit-1789466213","/_next/static/chunks/3q_n1v1z_hn1-.js?dpl=nogit-1789466213","/_next/static/chunks/14ynz93rnm6_r.js?dpl=nogit-1789466213","/_next/static/chunks/3zzkpug-bge13.js?dpl=nogit-1789466213","/_next/static/chunks/1ewppg55gz8ao.js?dpl=nogit-1789466213","/_next/static/chunks/2sghpash7yifv.js?dpl=nogit-1789466213","/_next/static/chunks/1r2o7os-xpck9.js?dpl=nogit-1789466213","/_next/static/chunks/3xskuweujaw05.js?dpl=nogit-1789466213","/_next/static/chunks/2y341-r8ncizo.js?dpl=nogit-1789466213","/_next/static/chunks/30dopbgd1dkf9.js?dpl=nogit-1789466213","/_next/static/chunks/3tn81b-gro1f1.js?dpl=nogit-1789466213","/_next/static/chunks/139ut7awapdtx.js?dpl=nogit-1789466213","/_next/static/chunks/0fx374hbtcqun.js?dpl=nogit-1789466213","/_next/static/chunks/3v9ybc6cyn2el.js?dpl=nogit-1789466213","/_next/static/chunks/1ksyn9m68446d.js?dpl=nogit-1789466213","/_next/static/chunks/28z7oiyzc2ir9.js?dpl=nogit-1789466213","/_next/static/chunks/0zg1om_k0m5zv.js?dpl=nogit-1789466213","/_next/static/chunks/1fn8bf7933b7y.js?dpl=nogit-1789466213","/_next/static/chunks/43ajf4ofaeu5y.js?dpl=nogit-1789466213","/_next/static/chunks/258b_3vm29gcy.js?dpl=nogit-1789466213","/_next/static/chunks/2dr--mw_fu802.js?dpl=nogit-1789466213","/_next/static/chunks/14k7sugwbq3i8.js?dpl=nogit-1789466213","/_next/static/chunks/0k562u9txct8-.js?dpl=nogit-1789466213"],"RecordResourceResume"] 3a:I[426249,["/_next/static/chunks/2c7cbuj0myjl0.js?dpl=nogit-1789466213","/_next/static/chunks/2juje0t3dlhbm.js?dpl=nogit-1789466213","/_next/static/chunks/19ijf68pshdcn.js?dpl=nogit-1789466213","/_next/static/chunks/04u0lp5-90-_i.js?dpl=nogit-1789466213","/_next/static/chunks/0i-8vweyyd2z1.js?dpl=nogit-1789466213","/_next/static/chunks/0br-rafekpauh.js?dpl=nogit-1789466213","/_next/static/chunks/0u008gt6051u7.js?dpl=nogit-1789466213","/_next/static/chunks/0zkcxwtbsqliz.js?dpl=nogit-1789466213","/_next/static/chunks/1wq0tmwxm0o03.js?dpl=nogit-1789466213","/_next/static/chunks/1bbv7o2-m-0ni.js?dpl=nogit-1789466213","/_next/static/chunks/3n6wyk-4-ou-2.js?dpl=nogit-1789466213","/_next/static/chunks/1h4wccj42vygd.js?dpl=nogit-1789466213","/_next/static/chunks/3jpl-42mbad5n.js?dpl=nogit-1789466213","/_next/static/chunks/1w_cokuayn2xb.js?dpl=nogit-1789466213","/_next/static/chunks/442oofikvq4ao.js?dpl=nogit-1789466213","/_next/static/chunks/2os7nxouq91eq.js?dpl=nogit-1789466213","/_next/static/chunks/3hxsq15e9l6ex.js?dpl=nogit-1789466213","/_next/static/chunks/3ht89vb0p09bm.js?dpl=nogit-1789466213","/_next/static/chunks/2qlkd_50jnb67.js?dpl=nogit-1789466213","/_next/static/chunks/3ml3vooxhiupb.js?dpl=nogit-1789466213","/_next/static/chunks/0ry5u_b6y0dbk.js?dpl=nogit-1789466213","/_next/static/chunks/0qfm0wv_qgrsj.js?dpl=nogit-1789466213","/_next/static/chunks/0gtr6g0fi5l0-.js?dpl=nogit-1789466213","/_next/static/chunks/14jjqpa8a5nws.js?dpl=nogit-1789466213","/_next/static/chunks/0i6x1920m4zk4.js?dpl=nogit-1789466213","/_next/static/chunks/1yts3efoejfo0.js?dpl=nogit-1789466213","/_next/static/chunks/04zr-wqafvwoa.js?dpl=nogit-1789466213","/_next/static/chunks/2id2jbelkqob6.js?dpl=nogit-1789466213","/_next/static/chunks/18kibg6v3ribz.js?dpl=nogit-1789466213","/_next/static/chunks/0k-y_4mcfub-p.js?dpl=nogit-1789466213","/_next/static/chunks/2_zyy4shki--m.js?dpl=nogit-1789466213","/_next/static/chunks/2ydqdr10ndcei.js?dpl=nogit-1789466213","/_next/static/chunks/0fia0hf7-7vfo.js?dpl=nogit-1789466213","/_next/static/chunks/43dq9llw-iyp8.js?dpl=nogit-1789466213","/_next/static/chunks/2dx-vzwyw9a3g.js?dpl=nogit-1789466213","/_next/static/chunks/0a_nnu2-xlcam.js?dpl=nogit-1789466213","/_next/static/chunks/12zn735ys1sga.js?dpl=nogit-1789466213","/_next/static/chunks/0llqhg6-bxn_b.js?dpl=nogit-1789466213","/_next/static/chunks/20xy3ur1zq5mk.js?dpl=nogit-1789466213","/_next/static/chunks/0fkz1t-1q4vvx.js?dpl=nogit-1789466213","/_next/static/chunks/2cu8u4002xlhk.js?dpl=nogit-1789466213","/_next/static/chunks/23dzzhwlmv97g.js?dpl=nogit-1789466213","/_next/static/chunks/2muh31f8m33t4.js?dpl=nogit-1789466213","/_next/static/chunks/0epkvyva8shyj.js?dpl=nogit-1789466213","/_next/static/chunks/03ql4_ajt89dj.js?dpl=nogit-1789466213","/_next/static/chunks/2b6zx8pzxydki.js?dpl=nogit-1789466213","/_next/static/chunks/0tapm_eeixiw6.js?dpl=nogit-1789466213","/_next/static/chunks/1tw7dwog40-j_.js?dpl=nogit-1789466213","/_next/static/chunks/3q_n1v1z_hn1-.js?dpl=nogit-1789466213","/_next/static/chunks/14ynz93rnm6_r.js?dpl=nogit-1789466213","/_next/static/chunks/3zzkpug-bge13.js?dpl=nogit-1789466213","/_next/static/chunks/1ewppg55gz8ao.js?dpl=nogit-1789466213","/_next/static/chunks/2sghpash7yifv.js?dpl=nogit-1789466213","/_next/static/chunks/1r2o7os-xpck9.js?dpl=nogit-1789466213","/_next/static/chunks/3xskuweujaw05.js?dpl=nogit-1789466213","/_next/static/chunks/2y341-r8ncizo.js?dpl=nogit-1789466213","/_next/static/chunks/30dopbgd1dkf9.js?dpl=nogit-1789466213","/_next/static/chunks/3tn81b-gro1f1.js?dpl=nogit-1789466213","/_next/static/chunks/139ut7awapdtx.js?dpl=nogit-1789466213","/_next/static/chunks/0fx374hbtcqun.js?dpl=nogit-1789466213","/_next/static/chunks/3v9ybc6cyn2el.js?dpl=nogit-1789466213","/_next/static/chunks/1ksyn9m68446d.js?dpl=nogit-1789466213","/_next/static/chunks/28z7oiyzc2ir9.js?dpl=nogit-1789466213","/_next/static/chunks/0zg1om_k0m5zv.js?dpl=nogit-1789466213","/_next/static/chunks/1fn8bf7933b7y.js?dpl=nogit-1789466213","/_next/static/chunks/43ajf4ofaeu5y.js?dpl=nogit-1789466213","/_next/static/chunks/258b_3vm29gcy.js?dpl=nogit-1789466213","/_next/static/chunks/2dr--mw_fu802.js?dpl=nogit-1789466213","/_next/static/chunks/14k7sugwbq3i8.js?dpl=nogit-1789466213","/_next/static/chunks/0k562u9txct8-.js?dpl=nogit-1789466213"],"TopicQuestionBank"] 25:["$","$L35",null,{"bundle":{"topicIds":["7fcd67bf-6119-477e-8aed-a01c33d6c3fc"],"paperIds":[],"stageIds":[],"stageTopicIds":[],"stageAssessmentIds":[],"topicNoteIds":[]},"children":[["$","$L36",null,{"feature":"question-bank"}],"$L37",["$","$L38",null,{"subjectCode":"a_level_maths","subjectName":"Maths","topicSlug":"a-level-pure-mathematics-5-4-solving-trigonometric-equations","topicTitle":"5.4 Solving Trigonometric Equations","sectionTitle":null,"activity":"worksheet","resourceTitle":"5.4 Solving Trigonometric Equations Questions","resourceHref":"/a-level/maths/edexcel/a-level-pure-mathematics-5-4-solving-trigonometric-equations/questions/q1","saveTrail":false}],["$","$L39",null,{"href":"/a-level/maths/edexcel/a-level-pure-mathematics-5-4-solving-trigonometric-equations/questions/q1","label":"5.4 Solving Trigonometric Equations","subjectCode":"a_level_maths","surface":"questions"}],["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"LearningResource\",\"name\":\"Question 5 – 5.4 Solving Trigonometric Equations - Questions\",\"description\":\"Practise 5.4 Solving Trigonometric Equations A Level Maths exam-style questions with worked solutions on MathsGenie.\",\"learningResourceType\":\"Practice questions\",\"educationalLevel\":\"A Level\",\"about\":{\"@type\":\"Thing\",\"name\":\"5.4 Solving Trigonometric Equations\"},\"url\":\"https://mathsgenie.co.uk/a-level/maths/edexcel/a-level-pure-mathematics-5-4-solving-trigonometric-equations/questions/q5\",\"inLanguage\":\"en-GB\",\"provider\":{\"@id\":\"https://mathsgenie.co.uk/#organization\"}}"}}],["$","div",null,{"className":"min-h-0 w-full flex-1 overflow-hidden bg-background","children":["$","$L3a",null,{"backHref":"/a-level/maths/edexcel/questions","basePath":"/a-level/maths/edexcel/a-level-pure-mathematics-5-4-solving-trigonometric-equations/questions","formulaUrl":"https://assets.mathsgenie.co.uk/board-resources/a-level-maths/edexcel/formula-book.pdf","initialIndex":4,"initialQuestionWindow":[{"id":"b88231c0-c4e7-4bf9-9b40-8dfa97f3e170","revisionId":"d7cdb22d-e858-4c7f-b471-30225081458b","questionNumber":null,"questionType":"LA","specification":"","difficulty":3,"difficultyBand":"medium","totalMarks":4,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"c2b79c85-1055-49ce-a3cb-a8c30ca4d444","label":null,"content":"The intensity of polarized light, \$ I \$, passing through a rotating filter is modeled by the equation \\[ I = I_0 \\cos^2 \\theta \\] where \$ I_0 \$ is the maximum intensity and \$ \\theta \$ is the angle of the filter in radians. In a specific experiment, the maximum intensity is measured as \$ I_0 = 12 \$ units. Determine all values of \$ \\theta \$ in the interval \$ 0 \\le \\theta \\le 2\\pi \$ for which the intensity \$ I \$ is exactly \$ 9 \$ units. Give your answers in terms of \$ \\pi \$.","markscheme":"- Substitutes $I = 9$ and $I_0 = 12$ and rearranges to obtain $\\cos^2 \\theta = \\frac{9}{12}$ or $0.75$ M1\n- Takes the square root of both sides to identify both $\\cos \\theta = \\frac{\\sqrt{3}}{2}$ and $\\cos \\theta = -\\frac{\\sqrt{3}}{2}$ M1\n- Finds at least two correct solutions for $\\theta$ from $\\{ \\frac{\\pi}{6}, \\frac{5\\pi}{6}, \\frac{7\\pi}{6}, \\frac{11\\pi}{6} \\}$ A1\n- Provides all four solutions: $\\theta = \\frac{\\pi}{6}$, $\\theta = \\frac{5\\pi}{6}$, $\\theta = \\frac{7\\pi}{6}$, and $\\theta = \\frac{11\\pi}{6}$ A1","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"algebraic_unit_circle","label":"Algebraic using Unit Circle","steps":[{"type":"text","order":0,"title":"Substitute and rearrange","content":"First, we substitute the given values into our intensity formula. We are given the maximum intensity $I_0 = 12$ and we want to find the angle $\theta$ where the intensity is $I = 9$.\n\nSubstitute these into the equation:\n$$9 = 12 \\cos^2 \\theta$$\n\nNow, divide both sides by $12$ to isolate the squared trigonometric term:\n$$\\cos^2 \\theta = \\frac{9}{12}$$\n\nSimplifying the fraction gives:\n$$\\cos^2 \\theta = \\frac{3}{4}$$","highlights":[{"text":"maximum intensity is measured as \$ I_0 = 12 \$ units.","location":"specification"},{"text":"intensity \$ I \$ is exactly \$ 9 \$ units.","location":"specification"}]},{"type":"text","order":1,"title":"Solve for cos θ","content":"To solve for $\\cos \\theta$, we take the square root of both sides. Remember that taking a square root introduces both positive and negative possibilities:\n\n$$\\cos \\theta = \\pm \\sqrt{\\frac{3}{4}}$$\n\nUsing the properties of square roots, we can simplify this to:\n$$\\cos \\theta = \\pm \\frac{\\sqrt{3}}{2}$$"},{"type":"text","order":2,"title":"Find the reference angle","content":"To find our solutions on the unit circle, we first identify the first-quadrant reference angle $\\theta_{\\text{ref}}$ where:\n$$\\cos \\theta_{\\text{ref}} = \\frac{\\sqrt{3}}{2}$$\n\nFrom our standard exact values, we know:\n$$\\theta_{\\text{ref}} = \\frac{\\pi}{6}$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"Using the symmetry of the unit circle, the equation $\\cos \\theta = \\pm\\frac{\\sqrt{3}}{2}$ has four solutions within the interval $0 \\le \\theta \\le 2\\pi$, corresponding to one angle in each quadrant:"},{"latex":"$$\\text{Quadrant I: } \\htmlData{anchor=q1}{\\theta = \\frac{\\pi}{6}}$"},{"latex":"$$\\text{Quadrant II: } \\htmlData{anchor=q2}{\\theta} = \\pi - \\frac{\\pi}{6} = \\htmlData{anchor=ans2}{\\frac{5\\pi}{6}}$"},{"latex":"$$\\text{Quadrant III: } \\htmlData{anchor=q3}{\\theta} = \\pi + \\frac{\\pi}{6} = \\htmlData{anchor=ans3}{\\frac{7\\pi}{6}}$"},{"latex":"$$\\text{Quadrant IV: } \\htmlData{anchor=q4}{\\theta} = 2\\pi - \\frac{\\pi}{6} = \\htmlData{anchor=ans4}{\\frac{11\\pi}{6}}$"}],"highlights":[{"color":"sky","style":"box","anchor":"q1"},{"color":"amber","style":"box","anchor":"ans2"},{"color":"purple","style":"box","anchor":"ans3"},{"color":"teal","style":"box","anchor":"ans4"}],"connections":[]},"order":3,"title":"Apply unit circle symmetry","description":"The positive value $\\cos \\theta = \\frac{\\sqrt{3}}{2}$ gives the solutions in Quadrants I and IV. The negative value $\\cos \\theta = -\\frac{\\sqrt{3}}{2}$ gives the solutions in Quadrants II and III."},{"type":"text","order":4,"title":"State the final solutions","content":"All four solutions lie within the required domain $0 \\le \\theta \\le 2\\pi$.\n\nThus, the values of $\\theta$ are:\n$$\\theta = \\frac{\\pi}{6}, \\quad \\theta = \\frac{5\\pi}{6}, \\quad \\theta = \\frac{7\\pi}{6}, \\quad \\text{and} \\quad \\theta = \\frac{11\\pi}{6}$$","highlights":[{"text":"0 \\le \\theta \\le 2\\pi","location":"specification"}]}]},{"id":"double_angle_identity","label":"Using Double-Angle Identity","steps":[{"type":"text","order":0,"title":"Substitute the given values","content":"First, we substitute the known values from the problem into our intensity model. We are given the maximum intensity $I_0 = 12$ units and we want to find when the intensity $I$ is exactly $9$ units:\n\n$$9 = 12 \\cos^2 \\theta$$ \n\nDividing both sides by $12$ simplifies to:\n\n$$\\cos^2 \\theta = \\frac{9}{12} = \\frac{3}{4}$$","highlights":[{"text":"maximum intensity is measured as \$ I_0 = 12 \$ units","location":"specification"},{"text":"intensity \$ I \$ is exactly \$ 9 \$ units","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=cos-sq}{\\color{@sky}{\\cos^2 \\theta}} = \\htmlData{anchor=double-id}{\\color{@sky}{\\frac{1 + \\cos(2\\theta)}{2}}}$"},{"latex":"$$\\htmlData{anchor=substitute}{\\color{@sky}{\\frac{1 + \\cos(2\\theta)}{2}}} = \\frac{3}{4}$"}],"highlights":[{"color":"sky","style":"box","anchor":"cos-sq"},{"color":"sky","style":"box","anchor":"double-id"}],"connections":[{"to":"substitute","from":"double-id","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Apply double-angle identity","description":"We rewrite $\\cos^2 \\theta$ using the double-angle identity: $$\\cos^2 \\theta = \\frac{1 + \\cos(2\\theta)}{2}$$ This transforms our quadratic equation into a linear trigonometric equation."},{"type":"text","order":2,"title":"Rearrange for cos(2θ)","content":"Now, we isolate the $\\cos(2\\theta)$ term. Multiplying both sides of the equation by $2$ gives:\n\n$$1 + \\cos(2\\theta) = 2 \\times \\frac{3}{4}$$\n\n$$1 + \\cos(2\\theta) = \\frac{3}{2}$$\n\nSubtracting $1$ from both sides yields:\n\n$$\\cos(2\\theta) = \\frac{3}{2} - 1$$\n\n$$\\cos(2\\theta) = \\frac{1}{2}$$"},{"type":"text","order":3,"title":"Adjust the domain","content":"The original interval for $\\theta$ is:\n\n$$0 \\le \\theta \\le 2\\pi$$\n\nSince our equation is in terms of the angle $2\\theta$, we must adjust our search interval by multiplying the entire inequality by $2$:\n\n$$0 \\le 2\\theta \\le 4\\pi$$","highlights":[{"text":"interval \$ 0 \\le \\theta \\le 2\\pi \$","location":"specification"}]},{"type":"text","order":4,"title":"Solve for 2θ","content":"We solve $\\cos(2\\theta) = \n\\frac{1}{2}$ within the interval $[0, 4\\pi]$.\n\nThe principal solution (in the first quadrant) is:\n\n$$2\\theta = \\frac{\\pi}{3}$$\n\nSince cosine is also positive in the fourth quadrant, our next solution within $[0, 2\\pi]$ is:\n\n$$2\\theta = 2\\pi - \\frac{\\pi}{3} = \\frac{5\n\\pi}{3}$$\n\nTo find the solutions in the second rotation $[2\\pi, 4\\pi]$, we add $2\\pi$ to each of our first two solutions:\n\n$$2\\theta = \\frac{\\pi}{3} + 2\\pi = \\frac{7\\pi}{3}$$\n\n$$2\\theta = \\frac{5\\pi}{3} + 2\\pi = \\frac{11\\pi}{3}$$"},{"type":"text","order":5,"title":"Determine values of θ","content":"To find the final values of $\\theta$, we divide each value of $2\\theta$ by $2$:\n\n$$\\theta = \\frac{\\pi}{6}, \\quad \\theta = \\frac{5\\pi}{6}, \\quad \\theta = \\frac{7\\pi}{6}, \\quad \\theta = \\frac{11\\pi}{6}$$\n\nAll four solutions lie within our specified domain of $0 \\le \\theta \\le 2\\pi$."}]}],"generatedAt":"2026-06-11T03:44:06.788Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"c2b79c85-1055-49ce-a3cb-a8c30ca4d444:algebraic_unit_circle","label":"Algebraic using Unit Circle","steps":[{"type":"text","order":0,"title":"Substitute and rearrange","content":"First, we substitute the given values into our intensity formula. We are given the maximum intensity $I_0 = 12$ and we want to find the angle $\theta$ where the intensity is $I = 9$.\n\nSubstitute these into the equation:\n$$9 = 12 \\cos^2 \\theta$$\n\nNow, divide both sides by $12$ to isolate the squared trigonometric term:\n$$\\cos^2 \\theta = \\frac{9}{12}$$\n\nSimplifying the fraction gives:\n$$\\cos^2 \\theta = \\frac{3}{4}$$","highlights":[{"text":"maximum intensity is measured as \$ I_0 = 12 \$ units.","location":"specification"},{"text":"intensity \$ I \$ is exactly \$ 9 \$ units.","location":"specification"}]},{"type":"text","order":1,"title":"Solve for cos θ","content":"To solve for $\\cos \\theta$, we take the square root of both sides. Remember that taking a square root introduces both positive and negative possibilities:\n\n$$\\cos \\theta = \\pm \\sqrt{\\frac{3}{4}}$$\n\nUsing the properties of square roots, we can simplify this to:\n$$\\cos \\theta = \\pm \\frac{\\sqrt{3}}{2}$$"},{"type":"text","order":2,"title":"Find the reference angle","content":"To find our solutions on the unit circle, we first identify the first-quadrant reference angle $\\theta_{\\text{ref}}$ where:\n$$\\cos \\theta_{\\text{ref}} = \\frac{\\sqrt{3}}{2}$$\n\nFrom our standard exact values, we know:\n$$\\theta_{\\text{ref}} = \\frac{\\pi}{6}$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"Using the symmetry of the unit circle, the equation $\\cos \\theta = \\pm\\frac{\\sqrt{3}}{2}$ has four solutions within the interval $0 \\le \\theta \\le 2\\pi$, corresponding to one angle in each quadrant:"},{"latex":"$$\\text{Quadrant I: } \\htmlData{anchor=q1}{\\theta = \\frac{\\pi}{6}}$"},{"latex":"$$\\text{Quadrant II: } \\htmlData{anchor=q2}{\\theta} = \\pi - \\frac{\\pi}{6} = \\htmlData{anchor=ans2}{\\frac{5\\pi}{6}}$"},{"latex":"$$\\text{Quadrant III: } \\htmlData{anchor=q3}{\\theta} = \\pi + \\frac{\\pi}{6} = \\htmlData{anchor=ans3}{\\frac{7\\pi}{6}}$"},{"latex":"$$\\text{Quadrant IV: } \\htmlData{anchor=q4}{\\theta} = 2\\pi - \\frac{\\pi}{6} = \\htmlData{anchor=ans4}{\\frac{11\\pi}{6}}$"}],"highlights":[{"color":"sky","style":"box","anchor":"q1"},{"color":"amber","style":"box","anchor":"ans2"},{"color":"purple","style":"box","anchor":"ans3"},{"color":"teal","style":"box","anchor":"ans4"}],"connections":[]},"order":3,"title":"Apply unit circle symmetry","description":"The positive value $\\cos \\theta = \\frac{\\sqrt{3}}{2}$ gives the solutions in Quadrants I and IV. The negative value $\\cos \\theta = -\\frac{\\sqrt{3}}{2}$ gives the solutions in Quadrants II and III."},{"type":"text","order":4,"title":"State the final solutions","content":"All four solutions lie within the required domain $0 \\le \\theta \\le 2\\pi$.\n\nThus, the values of $\\theta$ are:\n$$\\theta = \\frac{\\pi}{6}, \\quad \\theta = \\frac{5\\pi}{6}, \\quad \\theta = \\frac{7\\pi}{6}, \\quad \\text{and} \\quad \\theta = \\frac{11\\pi}{6}$$","highlights":[{"text":"0 \\le \\theta \\le 2\\pi","location":"specification"}]}]},{"id":"c2b79c85-1055-49ce-a3cb-a8c30ca4d444:double_angle_identity","label":"Using Double-Angle Identity","steps":[{"type":"text","order":0,"title":"Substitute the given values","content":"First, we substitute the known values from the problem into our intensity model. We are given the maximum intensity $I_0 = 12$ units and we want to find when the intensity $I$ is exactly $9$ units:\n\n$$9 = 12 \\cos^2 \\theta$$ \n\nDividing both sides by $12$ simplifies to:\n\n$$\\cos^2 \\theta = \\frac{9}{12} = \\frac{3}{4}$$","highlights":[{"text":"maximum intensity is measured as \$ I_0 = 12 \$ units","location":"specification"},{"text":"intensity \$ I \$ is exactly \$ 9 \$ units","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=cos-sq}{\\color{@sky}{\\cos^2 \\theta}} = \\htmlData{anchor=double-id}{\\color{@sky}{\\frac{1 + \\cos(2\\theta)}{2}}}$"},{"latex":"$$\\htmlData{anchor=substitute}{\\color{@sky}{\\frac{1 + \\cos(2\\theta)}{2}}} = \\frac{3}{4}$"}],"highlights":[{"color":"sky","style":"box","anchor":"cos-sq"},{"color":"sky","style":"box","anchor":"double-id"}],"connections":[{"to":"substitute","from":"double-id","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Apply double-angle identity","description":"We rewrite $\\cos^2 \\theta$ using the double-angle identity: $$\\cos^2 \\theta = \\frac{1 + \\cos(2\\theta)}{2}$$ This transforms our quadratic equation into a linear trigonometric equation."},{"type":"text","order":2,"title":"Rearrange for cos(2θ)","content":"Now, we isolate the $\\cos(2\\theta)$ term. Multiplying both sides of the equation by $2$ gives:\n\n$$1 + \\cos(2\\theta) = 2 \\times \\frac{3}{4}$$\n\n$$1 + \\cos(2\\theta) = \\frac{3}{2}$$\n\nSubtracting $1$ from both sides yields:\n\n$$\\cos(2\\theta) = \\frac{3}{2} - 1$$\n\n$$\\cos(2\\theta) = \\frac{1}{2}$$"},{"type":"text","order":3,"title":"Adjust the domain","content":"The original interval for $\\theta$ is:\n\n$$0 \\le \\theta \\le 2\\pi$$\n\nSince our equation is in terms of the angle $2\\theta$, we must adjust our search interval by multiplying the entire inequality by $2$:\n\n$$0 \\le 2\\theta \\le 4\\pi$$","highlights":[{"text":"interval \$ 0 \\le \\theta \\le 2\\pi \$","location":"specification"}]},{"type":"text","order":4,"title":"Solve for 2θ","content":"We solve $\\cos(2\\theta) = \n\\frac{1}{2}$ within the interval $[0, 4\\pi]$.\n\nThe principal solution (in the first quadrant) is:\n\n$$2\\theta = \\frac{\\pi}{3}$$\n\nSince cosine is also positive in the fourth quadrant, our next solution within $[0, 2\\pi]$ is:\n\n$$2\\theta = 2\\pi - \\frac{\\pi}{3} = \\frac{5\n\\pi}{3}$$\n\nTo find the solutions in the second rotation $[2\\pi, 4\\pi]$, we add $2\\pi$ to each of our first two solutions:\n\n$$2\\theta = \\frac{\\pi}{3} + 2\\pi = \\frac{7\\pi}{3}$$\n\n$$2\\theta = \\frac{5\\pi}{3} + 2\\pi = \\frac{11\\pi}{3}$$"},{"type":"text","order":5,"title":"Determine values of θ","content":"To find the final values of $\\theta$, we divide each value of $2\\theta$ by $2$:\n\n$$\\theta = \\frac{\\pi}{6}, \\quad \\theta = \\frac{5\\pi}{6}, \\quad \\theta = \\frac{7\\pi}{6}, \\quad \\theta = \\frac{11\\pi}{6}$$\n\nAll four solutions lie within our specified domain of $0 \\le \\theta \\le 2\\pi$."}]}],"generatedAt":"2026-06-11T03:44:07.880Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"db08208d-c428-4a8d-a4fb-84fbf9f58ef6","revisionId":"d7e9b9d5-a6f6-4e86-88e8-ceb179aac570","questionNumber":null,"questionType":"LA","specification":"","difficulty":3,"difficultyBand":"medium","totalMarks":4,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"d1c58540-8f26-45a0-ab7a-8856331d7374","label":null,"content":"Solve the equation \\[ 2\\cos^2 x = \\frac{1}{2} \\] for \$ 0^\\circ \\le x \\le 360^\\circ \$.","markscheme":"- Rearranges the equation to $\\cos^2 x = \\frac{1}{4}$ M1\n- Takes the square root of both sides to obtain $\\cos x = \\pm \\frac{1}{2}$ M1\n- Finds solutions for $\\cos x = \\frac{1}{2}$ as $x = 60^\\circ$ and $x = 360^\\circ - 60^\\circ = 300^\\circ$ A1\n- Finds solutions for $\\cos x = -\\frac{1}{2}$ as $x = 180^\\circ - 60^\\circ = 120^\\circ$ and $x = 180^\\circ + 60^\\circ = 240^\\circ$ A1","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"square_root","label":"Direct Square Root","steps":[{"type":"text","order":0,"title":"Isolate the squared term","content":"To solve the equation, we first want to get the $\\cos^2 x$ term by itself.\n\nWe do this by dividing both sides of the equation by $2$:\n\n$$\\begin{aligned} 2\\cos^2 x &= \\frac{1}{2} \\\\ \\cos^2 x &= \\frac{1}{4} \\end{aligned}$$\n\nThis step earns you the first mark (**M1**)."},{"type":"text","order":1,"title":"Take the square root","content":"Next, take the square root of both sides. Remember to include both the positive and negative square roots!\n\n$$\\cos x = \\pm \\sqrt{\\frac{1}{4}}$$\n\nThis gives us two linear trigonometric equations to solve:\n\n$$\\cos x = \\frac{1}{2} \\quad \\text{or} \\quad \\cos x = -\\frac{1}{2}$$\n\nThis step earns you the second mark (**M1**)."},{"type":"text","order":2,"title":"Solve for positive case","content":"Let's find the solutions for the positive case, $\\cos x = \\frac{1}{2}$, within the interval $0^\\circ \\le x \\le 360^\\circ$.\n\nThe principal value is:\n$$x = \\arccos\\left(\\frac{1}{2}\\right) = 60^\\circ$$\n\nUsing the symmetry of the cosine graph, the other solution in the interval is:\n$$x = 360^\\circ - 60^\\circ = 300^\\circ$$\n\nThis gives us the first two solutions (**A1**):\n$$x = 60^\\circ, \\; 300^\\circ$$"},{"type":"text","order":3,"title":"Solve for negative case","content":"Now we find the solutions for the negative case, $\\cos x = -\\frac{1}{2}$, within the interval $0^\\circ \\le x \\le 360^\\circ$.\n\nUsing our reference angle of $60^\\circ$, cosine is negative in the second and third quadrants:\n- **Quadrant II:** $x = 180^\\circ - 60^\\circ = 120^\\circ$\n- **Quadrant III:** $x = 180^\\circ + 60^\\circ = 240^\\circ$\n\nThis gives us the remaining two solutions (**A1**):\n$$x = 120^\\circ, \\; 240^\\circ$$"},{"type":"text","order":4,"title":"State final solutions","content":"Combining the results from both cases, we get the complete set of solutions for $0^\\circ \\le x \\le 360^\\circ$:\n\n$$x = 60^\\circ, \\; 120^\\circ, \\; 240^\\circ, \\; 300^\\circ$$","calculator":[{"gdc":{"expression":"y = 2*cos(x)^2","instructions":"1. Put the calculator in **Degree** mode.\n2. Go to the Graphing menu.\n3. Enter the equations $y1 = 2\\cos(x)^2$ and $y2 = 0.5$.\n4. Set the viewing window $x$ from $0$ to $360$ and $y$ from $-0.5$ to $2.5$.\n5. Use the **G-Solve** (or Intersect) tool to find the four points where the curves cross."},"ti84":{"expression":"2 * (cos(X))^2 = 0.5","instructions":"1. Press [MODE] and ensure your calculator is set to **DEGREE**.\n2. Press [Y=] and enter:\n - `Y1 = 2 * (cos(X))^2`\n - `Y2 = 0.5`\n3. Press [WINDOW] and set:\n - `Xmin = 0`\n - `Xmax = 360`\n - `Xscl = 30`\n - `Ymin = -0.5`\n - `Ymax = 2.5`\n4. Press [GRAPH].\n5. To find the intersections, press [2nd] [CALC] and select `5: intersect`. Move the cursor near each intersection point and press [ENTER] three times to find the values: $60^\\circ$, $120^\\circ$, $240^\\circ$, and $300^\\circ$."},"desmos":{"expression":"y = 2\\cos^2(x)","instructions":"1. Click the wrench icon (Settings) and select **Degrees**.\n2. Set the X-axis range from `0` to `360`.\n3. Type `y = 2(cos(x))^2` in the first box.\n4. Type `y = 0.5` in the second box.\n5. Click on the gray intersection points along the lines to read the coordinates: $(60, 0.5)$, $(120, 0.5)$, $(240, 0.5)$, and $(300, 0.5)$."},"description":"You can verify your answers graphically by finding where the function $y = 2\\cos^2(x)$ intersects the line $y = 0.5$."}]}]},{"id":"double_angle","label":"Double Angle Identity","steps":[{"type":"text","order":0,"title":"Recall the double-angle identity","content":"To solve this using the double-angle identity approach, we use the identity:\n\n$$2\\cos^2 x = 1 + \\cos(2x)$$\n\nThis allows us to rewrite the quadratic expression in terms of a single cosine function of double the angle."},{"type":"text","order":1,"title":"Substitute and simplify","content":"Substituting the identity into the original equation, we get:\n\n$$1 + \\cos(2x) = \\frac{1}{2}$$\n\nNow, subtract $1$ from both sides to solve for $\\cos(2x)$:\n\n$$\\cos(2x) = -\\frac{1}{2}$$"},{"type":"text","order":2,"title":"Adjust the domain","content":"Our original domain is for $x$:\n\n$$0^\\circ \\le x \\le 360^\\circ$$\n\nSince our equation is in terms of $2x$, we must adjust the domain by multiplying all parts by $2$:\n\n$$0^\\circ \\le 2x \\le 720^\\circ$$\n\nWe need to find all solutions for $2x$ within this larger interval."},{"type":"text","order":3,"title":"Solve for the angle 2x","content":"We look for angles where $\\cos(2x) = -\\frac{1}{2}$ within the interval $[0^\\circ, 720^\\circ]$.\n\nThe principal solution is:\n\n$$2x = \\arccos\\left(-\\frac{1}{2}\\right) = 120^\\circ$$\n\nUsing the symmetry of the cosine graph:\n\n$$2x = 360^\\circ - 120^\\circ = 240^\\circ$$\n\nTo find the solutions in the second cycle ($360^\\circ \\le 2x \\le 720^\\circ$), we add $360^\\circ$ to our first-cycle solutions:\n\n$$2x = 120^\\circ + 360^\\circ = 480^\\circ$$\n\n$$2x = 240^\\circ + 360^\\circ = 600^\\circ$$"},{"type":"text","order":4,"title":"Find the values of x","content":"Finally, divide each value of $2x$ by $2$ to find the values of $x$:\n\n$$\\begin{aligned} x &= \\frac{120^\\circ}{2} = 60^\\circ \\\\ x &= \\frac{240^\\circ}{2} = 120^\\circ \\\\ x &= \\frac{480^\\circ}{2} = 240^\\circ \\\\ x &= \\frac{600^\\circ}{2} = 300^\\circ \\end{aligned}$$\n\nThus, the solutions in the interval $0^\\circ \\le x \\le 360^\\circ$ are $x = 60^\\circ, 120^\\circ, 240^\\circ, 300^\\circ$."}]}],"generatedAt":"2026-06-09T08:32:59.469Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"d1c58540-8f26-45a0-ab7a-8856331d7374:square_root","label":"Direct Square Root","steps":[{"type":"text","order":0,"title":"Isolate the squared term","content":"To solve the equation, we first want to get the $\\cos^2 x$ term by itself.\n\nWe do this by dividing both sides of the equation by $2$:\n\n$$\\begin{aligned} 2\\cos^2 x &= \\frac{1}{2} \\\\ \\cos^2 x &= \\frac{1}{4} \\end{aligned}$$\n\nThis step earns you the first mark (**M1**)."},{"type":"text","order":1,"title":"Take the square root","content":"Next, take the square root of both sides. Remember to include both the positive and negative square roots!\n\n$$\\cos x = \\pm \\sqrt{\\frac{1}{4}}$$\n\nThis gives us two linear trigonometric equations to solve:\n\n$$\\cos x = \\frac{1}{2} \\quad \\text{or} \\quad \\cos x = -\\frac{1}{2}$$\n\nThis step earns you the second mark (**M1**)."},{"type":"text","order":2,"title":"Solve for positive case","content":"Let's find the solutions for the positive case, $\\cos x = \\frac{1}{2}$, within the interval $0^\\circ \\le x \\le 360^\\circ$.\n\nThe principal value is:\n$$x = \\arccos\\left(\\frac{1}{2}\\right) = 60^\\circ$$\n\nUsing the symmetry of the cosine graph, the other solution in the interval is:\n$$x = 360^\\circ - 60^\\circ = 300^\\circ$$\n\nThis gives us the first two solutions (**A1**):\n$$x = 60^\\circ, \\; 300^\\circ$$"},{"type":"text","order":3,"title":"Solve for negative case","content":"Now we find the solutions for the negative case, $\\cos x = -\\frac{1}{2}$, within the interval $0^\\circ \\le x \\le 360^\\circ$.\n\nUsing our reference angle of $60^\\circ$, cosine is negative in the second and third quadrants:\n- **Quadrant II:** $x = 180^\\circ - 60^\\circ = 120^\\circ$\n- **Quadrant III:** $x = 180^\\circ + 60^\\circ = 240^\\circ$\n\nThis gives us the remaining two solutions (**A1**):\n$$x = 120^\\circ, \\; 240^\\circ$$"},{"type":"text","order":4,"title":"State final solutions","content":"Combining the results from both cases, we get the complete set of solutions for $0^\\circ \\le x \\le 360^\\circ$:\n\n$$x = 60^\\circ, \\; 120^\\circ, \\; 240^\\circ, \\; 300^\\circ$$","calculator":[{"gdc":{"expression":"y = 2*cos(x)^2","instructions":"1. Put the calculator in **Degree** mode.\n2. Go to the Graphing menu.\n3. Enter the equations $y1 = 2\\cos(x)^2$ and $y2 = 0.5$.\n4. Set the viewing window $x$ from $0$ to $360$ and $y$ from $-0.5$ to $2.5$.\n5. Use the **G-Solve** (or Intersect) tool to find the four points where the curves cross."},"ti84":{"expression":"2 * (cos(X))^2 = 0.5","instructions":"1. Press [MODE] and ensure your calculator is set to **DEGREE**.\n2. Press [Y=] and enter:\n - `Y1 = 2 * (cos(X))^2`\n - `Y2 = 0.5`\n3. Press [WINDOW] and set:\n - `Xmin = 0`\n - `Xmax = 360`\n - `Xscl = 30`\n - `Ymin = -0.5`\n - `Ymax = 2.5`\n4. Press [GRAPH].\n5. To find the intersections, press [2nd] [CALC] and select `5: intersect`. Move the cursor near each intersection point and press [ENTER] three times to find the values: $60^\\circ$, $120^\\circ$, $240^\\circ$, and $300^\\circ$."},"desmos":{"expression":"y = 2\\cos^2(x)","instructions":"1. Click the wrench icon (Settings) and select **Degrees**.\n2. Set the X-axis range from `0` to `360`.\n3. Type `y = 2(cos(x))^2` in the first box.\n4. Type `y = 0.5` in the second box.\n5. Click on the gray intersection points along the lines to read the coordinates: $(60, 0.5)$, $(120, 0.5)$, $(240, 0.5)$, and $(300, 0.5)$."},"description":"You can verify your answers graphically by finding where the function $y = 2\\cos^2(x)$ intersects the line $y = 0.5$."}]}]},{"id":"d1c58540-8f26-45a0-ab7a-8856331d7374:double_angle","label":"Double Angle Identity","steps":[{"type":"text","order":0,"title":"Recall the double-angle identity","content":"To solve this using the double-angle identity approach, we use the identity:\n\n$$2\\cos^2 x = 1 + \\cos(2x)$$\n\nThis allows us to rewrite the quadratic expression in terms of a single cosine function of double the angle."},{"type":"text","order":1,"title":"Substitute and simplify","content":"Substituting the identity into the original equation, we get:\n\n$$1 + \\cos(2x) = \\frac{1}{2}$$\n\nNow, subtract $1$ from both sides to solve for $\\cos(2x)$:\n\n$$\\cos(2x) = -\\frac{1}{2}$$"},{"type":"text","order":2,"title":"Adjust the domain","content":"Our original domain is for $x$:\n\n$$0^\\circ \\le x \\le 360^\\circ$$\n\nSince our equation is in terms of $2x$, we must adjust the domain by multiplying all parts by $2$:\n\n$$0^\\circ \\le 2x \\le 720^\\circ$$\n\nWe need to find all solutions for $2x$ within this larger interval."},{"type":"text","order":3,"title":"Solve for the angle 2x","content":"We look for angles where $\\cos(2x) = -\\frac{1}{2}$ within the interval $[0^\\circ, 720^\\circ]$.\n\nThe principal solution is:\n\n$$2x = \\arccos\\left(-\\frac{1}{2}\\right) = 120^\\circ$$\n\nUsing the symmetry of the cosine graph:\n\n$$2x = 360^\\circ - 120^\\circ = 240^\\circ$$\n\nTo find the solutions in the second cycle ($360^\\circ \\le 2x \\le 720^\\circ$), we add $360^\\circ$ to our first-cycle solutions:\n\n$$2x = 120^\\circ + 360^\\circ = 480^\\circ$$\n\n$$2x = 240^\\circ + 360^\\circ = 600^\\circ$$"},{"type":"text","order":4,"title":"Find the values of x","content":"Finally, divide each value of $2x$ by $2$ to find the values of $x$:\n\n$$\\begin{aligned} x &= \\frac{120^\\circ}{2} = 60^\\circ \\\\ x &= \\frac{240^\\circ}{2} = 120^\\circ \\\\ x &= \\frac{480^\\circ}{2} = 240^\\circ \\\\ x &= \\frac{600^\\circ}{2} = 300^\\circ \\end{aligned}$$\n\nThus, the solutions in the interval $0^\\circ \\le x \\le 360^\\circ$ are $x = 60^\\circ, 120^\\circ, 240^\\circ, 300^\\circ$."}]}],"generatedAt":"2026-06-09T08:33:11.242Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths","summaries":[{"id":"6e256c92-2764-44c4-bdb0-431f4684f0c5","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"867e1c4c-5bdb-4ccf-ac8a-f128bc99febe","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"8d82aca5-86df-43e2-873e-af06a2934430","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"b88231c0-c4e7-4bf9-9b40-8dfa97f3e170","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"db08208d-c428-4a8d-a4fb-84fbf9f58ef6","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true}],"title":"5.4 Solving Trigonometric Equations","topicId":"7fcd67bf-6119-477e-8aed-a01c33d6c3fc","topicSidebar":"$L3b","topicSlug":"a-level-pure-mathematics-5-4-solving-trigonometric-equations","topicTitle":"5.4 Solving Trigonometric Equations"}]}],"$L3c","$L3d"]}]