## Precalculus (6th Edition)

$\frac{\pi}{4}$, $\frac{2\pi}{3}$, $\frac{5\pi}{4}$, $\frac{5\pi}{3}$
$(\cot x-1)(\sqrt{3}\cot x+1)=0$ $\cot x-1=0$ 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 $\frac{5\pi}{3}$.