Answer
$x=\frac{\pi}{4}+n\pi$ and $x=arctan(-2)+n\pi$, where $n$ is an integer.
Work Step by Step
$sec^2x=1+tan^2x$
$sec^2x+tan~x=3$
$1+tan^2x+tan~x-3=0$
$tan^2x+tan~x-2=0~~$ ($tan~x=2~tan~x-tan~x$)
$tan^2x+2~tan~x-tan~x-2=0$
$tan~x(tan~x+2)-1(tan~x+2)=0$
$(tan~x-1)(tan~x+2)=0$
$tan~x-1=0$
$tan~x=1$
$x=arctan~1=\frac{\pi}{4}$
$tan~x+2=0$
$tan~x=-2$
$x=arctan(-2)$
The period of $tan~x$ is $\pi$. So, add multiples of $\pi$ to each solution to find the general solution:
$x=\frac{\pi}{4}+n\pi$ and $x=arctan(-2)+n\pi$, where $n$ is an integer.
