Show that the equation
3sin2θ−2cos2θ=1 3 \sin^2 \theta - 2 \cos^2 \theta = 1 3sin2θ−2cos2θ=1can be written as
5sin2θ=3. 5 \sin^2 \theta = 3. 5sin2θ=3.Hence solve, for 0∘≤θ<360∘0^\circ \leq \theta < 360^\circ0∘≤θ<360∘, the equation
3sin2θ−2cos2θ=1, 3 \sin^2 \theta - 2 \cos^2 \theta = 1, 3sin2θ−2cos2θ=1,giving your answer to 1 decimal place.
Practise Edexcel A Level Old Maths Trigonometry with exam-style questions for A Level Old Maths. 13 questions, matched to the Edexcel A Level Old Maths (9371) specification and written in Unit exams C1, C2, C3 and C4 plus two applied units (e.g. S1 and M1) style. Every question includes a full worked solution and mark scheme, so you can see where marks are awarded rather than just whether you got the answer right.