Answer
$\theta=arctan-3+n\pi$
$\theta=arctan~2+n\pi$
where $n$ is an integer.
Work Step by Step
$tan^2\theta+tan~\theta-6=0$
$tan^2\theta+3~tan~\theta-2~tan~\theta-6=0$
$tan~\theta(tan~\theta+3)-2(tan~\theta+3)=0$
$(tan~\theta+3)(tan~\theta-2)=0$
$tan~\theta=-3$ or $tan~\theta=2$
$tan~\theta=-3$
$\theta=arctan-3$
$tan~\theta=2$
$\theta=arctan~2$
The period of $tan~\theta$ is $\pi$. The general solutions are:
$\theta=arctan-3+n\pi$
$\theta=arctan~2+n\pi$
where $n$ is an integer.
