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1.7 Trigonometry

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Question 82
a.

Solve 6sin⁡2θ=cos⁡θ+46 \sin^2 \theta = \cos \theta + 46sin2θ=cosθ+4 giving all the solutions for the interval 0≤θ<360∘0 \leq \theta < 360^\circ0≤θ<360∘

[4]
b.

Hence, solve 6sin⁡22θ=cos⁡2θ+46 \sin^2 2\theta = \cos 2\theta + 46sin22θ=cos2θ+4 giving all the solutions for the interval 0≤θ<360∘0 \leq \theta < 360^\circ0≤θ<360∘

[2]

1.7 Trigonometry Questions

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