Answer
$x=\{\frac{\pi}{3},\frac{4\pi}{3},\frac{5\pi}{6},\frac{11\pi}{6}\}$
Work Step by Step
$tan(2x)=-\sqrt{3}$
$2x=tan^{-1}(-\sqrt{3})$
We know $tan(x)$ is negative in quardent $II$ and quardent $IV$
The period of the tangent function is $2\pi$
$2x=\pi-\frac{\pi}{3}\;\;\;\;or\;\;\;\;\;\;2x=2\pi+ \frac{2\pi}{3}\;\;\;\;or\;\;\;\;\;\;2x=2\pi - \frac{5\pi}{3}\;\;\;\;\;\;or\;\;\;\;2x=2\pi + \frac{7\pi}{3} $
$x=\frac{\pi}{3}\;\;\;\;or\;\;\;\;\;\;x=\frac{4\pi}{3}\;\;\;\;or\;\;\;\;\;\;x=\frac{5\pi}{6}\;\;\;\;\;\;\;\;or\;\;\;\;\;x=\frac{11\pi}{6}$
$x=\{\frac{\pi}{3},\frac{4\pi}{3},\frac{5\pi}{6},\frac{11\pi}{6}\}$
