Find the general solution of .

MATHEMATICS · Single correct · A Level · Mathematics · Trigonometry · Trigonometry

Problem

Find the general solution of

\sin 2\theta = \dfrac{\sqrt3}{2}

. (a)

\theta = \dfrac{\pi}{3}+n\pi \;\text{or}\; \theta = \dfrac{2\pi}{3}+n\pi,\; n\in\mathbb Z

(b)

\theta = \dfrac{\pi}{6}+n\pi \;\text{or}\; \theta = \dfrac{\pi}{3}+n\pi,\; n\in\mathbb Z

(c)

\theta = \dfrac{\pi}{6}+2n\pi \;\text{or}\; \theta = \dfrac{5\pi}{6}+2n\pi,\; n\in\mathbb Z

(d)

\theta = \dfrac{\pi}{12}+n\pi \;\text{or}\; \theta = \dfrac{5\pi}{12}+n\pi,\; n\in\mathbb Z