Answer
$\frac{\pi}{4}$, $\frac{2\pi}{3}$, $\frac{5\pi}{4}$, $\frac{5\pi}{3}$
Work Step by Step
$(\cot x-1)(\sqrt{3}\cot x+1)=0$
$\cot x-1=0$ or $\sqrt{3}\cot x+1=0$
If $\cot x-1=0$:
$\cot x=1$
$\tan x=\frac{1}{1}=1$
The only solutions in $[0, 2\pi)$ are $\frac{\pi}{4}$ and $\frac{5\pi}{4}$.
If $\sqrt{3}\cot x+1=0$:
$\sqrt{3}\cot x=-1$
$\cot x=-\frac{1}{\sqrt{3}}$
$\tan x=-\sqrt{3}$
The only solutions in $[0, 2\pi)$ are $\frac{2\pi}{3}$ and $\frac{5\pi}{3}$.
