Answer
$x=arctan~2+n\pi$ and $x=-\frac{\pi}{4}+n\pi$, where $n$ is an integer.
Work Step by Step
Use: $-tan~x=tan~x-2~tan~x$
$tan^2x-tan~x-2=0~~$
$tan^2x+tan~x-2~tan~x-2=0$
$tan~x(tan~x+1)-2(tan~x+1)=0$
$(tan~x-2)(tan~x+1)=0$
$tan~x-2=0$
$tan~x=2$
$x=arctan~2$
$tan~x+1=0$
$tan~x=-1$
$x=arctan(-1)=-\frac{\pi}{4}$
The period of $tan~x$ is $\pi$. So, add multiples of $\pi$ to each solution to find the general solution:
$x=arctan~2+n\pi$ and $x=-\frac{\pi}{4}+n\pi$, where $n$ is an integer.
