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