Answer
$x=arctan(\frac{1}{3})+n\pi$ and $x=arctan(-\frac{1}{3})+n\pi$, where $n$ is an integer.
Work Step by Step
$cot^2x-9=0$
$cot^2x=9$
$cot~x=±3$
$tan~x=±\frac{1}{3}$
$x=arctan(\frac{1}{3})$
$x=arctan(-\frac{1}{3})$
The period of $tan~x$ is $\pi$. So, add multiples of $\pi$ to each solution to find the general solution:
$x=arctan(\frac{1}{3})+n\pi$ and $x=arctan(-\frac{1}{3})+n\pi$, where $n$ is an integer.
