Answer
$x=\frac{\pi}{4}+n\pi$ and $x=arctan(5)+n\pi$, where $n$ is an integer.
Work Step by Step
$sec^2x=1+tan^2x$
$sec^2x-6~tan~x=-4$
$1+tan^2x-6~tan~x+4=0$
$tan^2x-6~tan~x+5=0~~$ ($-6~tan~x=-5~tan~x-tan~x$)
$tan^2x-5~tan~x-tan~x+5=0$
$tan~x(tan~x-5)-1(tan~x-5)=0$
$(tan~x-1)(tan~x-5)=0$
$tan~x-1=0$
$tan~x=1$
$x=arctan~1=\frac{\pi}{4}$
$tan~x-5=0$
$tan~x=5$
$x=arctan(5)$
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 $x=arctan(5)+n\pi$, where $n$ is an integer.
