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MATHEMATICS · Single correct · AP · AP Precalculus · Trigonometric and Polar Functions · Trigonometric and Polar Functions

Problem

Solve sin⁡2θ=cos⁡θ\sin 2\theta = \cos\theta for θ\theta in (0,2π)(0, 2\pi). (a) θ=π6,  π2,  5π6,  3π2\theta = \dfrac{\pi}{6},\;\dfrac{\pi}{2},\;\dfrac{5\pi}{6},\;\dfrac{3\pi}{2} (b) θ=π3,  π2,  π,  3π2\theta = \dfrac{\pi}{3},\;\dfrac{\pi}{2},\;\pi,\;\dfrac{3\pi}{2} (c) θ=π6,  π2,  5π6\theta = \dfrac{\pi}{6},\;\dfrac{\pi}{2},\;\dfrac{5\pi}{6} (d) θ=π6,  π2,  3π2\theta = \dfrac{\pi}{6},\;\dfrac{\pi}{2},\;\dfrac{3\pi}{2}

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