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