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Trigonometric Identities and Equations

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Question 11
a.

Show that

6cos⁡2θ+7sin⁡θ−81−2sin⁡θ≡3sin⁡θ−2 \frac{6 \cos^2 \theta + 7 \sin \theta - 8}{1 - 2 \sin \theta} \equiv 3 \sin \theta - 2 1−2sinθ6cos2θ+7sinθ−8​≡3sinθ−2
[4]
b.

Hence solve, for 0≤θ<360∘0 \leq \theta < 360^\circ0≤θ<360∘, the equation,

6cos⁡2θ+7sin⁡θ−81−2sin⁡θ=2cos⁡θ−2 \frac{6 \cos^2 \theta + 7 \sin \theta - 8}{1 - 2 \sin \theta} = 2 \cos \theta - 2 1−2sinθ6cos2θ+7sinθ−8​=2cosθ−2
[3]

Trigonometric Identities and Equations Questions

  1. AS Level
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