Answer
$x=0, \pi$
Work Step by Step
Use the double angle identity such as: $ \tan 2x= \dfrac{2 \tan x}{1-\tan^2 x}$
or, $ \tan x-\dfrac{2 \tan x}{1-\tan^2 x}=0$
$\implies \tan x-\tan^3 x-2 \tan x=0$
or, $- \tan x(\tan^2 x+1)=0$
This implies that $\tan x=0$ and $\tan^2 x+1=0$
Thus, we have $x=0, \pi$