Answer
$x=arctan-5+n\pi$
$x=-\frac{\pi}{4}+n\pi$
where $n$ is an integer.
Work Step by Step
$sec^2x+6~tan~x+4=0$
$1+tan^2x+6~tan~x+4=0$
$tan^2x+6~tan~x+5=0$
$tan^2x+tan~x+5~tan~x+5=0$
$tan~x(tan~x+1)+5(tan~x+1)=0$
$(tan~x+5)(tan~x+1)=0$
$tan~x=-5$ or $tan~x=-1$
$tan~x=-5$
$x=arctan-5$
$tan~x=-1$
$x=-\frac{\pi}{4}$
The period of $tan~x$ is $\pi$. The general solutions are:
$x=arctan-5+n\pi$
$x=-\frac{\pi}{4}+n\pi$
where $n$ is an integer.
