## Algebra 2 (1st Edition)

$x=0, \pi$
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 $x=0, \pi$