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1 Unit AS 1: Pure Mathematics

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Question 340

Consider 2sin⁡2x=7cos⁡x+52\sin^2x=7\cos x+52sin2x=7cosx+5.

a.

Show that it can be written as 2cos⁡2x+7cos⁡x+3=02\cos^2x+7\cos x+3=02cos2x+7cosx+3=0.

[2]
b.

Hence solve it for 0∘≤x<360∘0^\circ\leq x<360^\circ0∘≤x<360∘.

[3]
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1 Unit AS 1: Pure Mathematics Questions

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