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Trigonometry and Modelling

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Question 6
a.

Show that

cos⁡θ(4tan⁡θ+3tan⁡θ)≡sin⁡θ+3sin⁡θ \cos \theta \left( 4 \tan \theta + \frac{3}{\tan \theta} \right) \equiv \sin \theta + \frac{3}{\sin \theta} cosθ(4tanθ+tanθ3​)≡sinθ+sinθ3​

for θ≠nπ2\theta \neq \frac{n\pi}{2}θ=2nπ​.

[4]
b.

Hence solve, for 0<x<2π0 < x < 2\pi0<x<2π, the equation

cos⁡x(4tan⁡x+3tan⁡x)=6sin⁡x−1 \cos x \left( 4 \tan x + \frac{3}{\tan x} \right) = 6 \sin x - 1 cosx(4tanx+tanx3​)=6sinx−1

giving your answers to 3 significant figures.

[6]

Trigonometry and Modelling Questions

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