Answer
$x=arctan~3+n\pi$ and $x=arctan(-4)+n\pi$, where $n$ is an integer.
Work Step by Step
Use: $tan~x=4~tan~x-3~tan~x$
$tan^2x+tan~x-12=0~~$
$tan^2x+4~tan~x-3~tan~x-12=0$
$tan~x(tan~x+4)-3(tan~x+4)=0$
$(tan~x-3)(tan~x+4)=0$
$tan~x-3=0$
$tan~x=3$
$x=arctan~3$
$tan~x+4=0$
$tan~x=-4$
$x=arctan(-4)$
The period of $tan~x$ is $\pi$. So, add multiples of $\pi$ to each solution to find the general solution:
$x=arctan~3+n\pi$ and $x=arctan(-4)+n\pi$, where $n$ is an integer.
