## Precalculus (10th Edition)

$\theta=\dfrac{2\pi}{3}$
$\tan(\frac{\theta}{2})=\sqrt3$ $\frac{\theta}{2}=\frac{1}{3}\pi+k\pi$ (since the period of tangent is $\pi$) $\theta=\frac{2}{3}\pi+2k\pi$ Those solutions where $0\le\theta\lt2\pi$ are only: $\theta=\frac{2}{3}\pi+2(0)\pi=\frac{2}{3}\pi$