Mathematics

Problem

Solve

\tan 2\theta = \sqrt{3}

for

\theta \in (-\pi/2,\,\pi/2)

. a)

\theta = -\dfrac{\pi}{3},\;\dfrac{\pi}{3}

\theta = -\dfrac{\pi}{3},\;\dfrac{\pi}{6}

\theta = -\dfrac{\pi}{6},\;\dfrac{\pi}{3}

\theta = -\dfrac{\pi}{6},\;\dfrac{\pi}{6}