Answer
$x=\frac{\pi}{3}+n\pi$ and $x=\frac{2\pi}{3}+n\pi$, where $n$ is an integer.
Work Step by Step
$3~cot^2x-1=0$
$3~cot^2x=1$
$cot^2x=\frac{1}{3}$
$cot~x=±\frac{1}{\sqrt 3}=±\frac{1}{\sqrt 3}\frac{\sqrt 3}{\sqrt 3}=±\frac{\sqrt 3}{3}$
The period of $cot~x$ is $\pi$. The solutions in the interval: $[0,\pi)$ is:
$x=\frac{\pi}{3}$ and $x=\frac{2\pi}{3}$
Now, add multiples of $\pi$ to the solution:
$x=\frac{\pi}{3}+n\pi$ and $x=\frac{2\pi}{3}+n\pi$, where $n$ is an integer.
