Answer
$\frac{\pi}{3}$, $\frac{2\pi}{3}$, $\frac{3\pi}{4}$, $\frac{4\pi}{3}$, $\frac{5\pi}{3}$, $\frac{7\pi}{4}$
Work Step by Step
$\tan^3 x+\tan^2 x-3\tan x-3=0$
$\tan^2 x(\tan x+1)-3(\tan x+1)=0$
$(\tan^2x-3)(\tan x+1)=0$
$(\tan x+\sqrt{3})(\tan x-\sqrt{3})(\tan x+1)=0$
If $\tan x+\sqrt{3}=0$, then $\tan x=-\sqrt{3}$, and the only solutions in $[0, 2\pi)$ are $\frac{2\pi}{3}$ and $\frac{5\pi}{3}$.
If $\tan x-\sqrt{3}=0$, then $\tan x=\sqrt{3}$, and the only solutions in $[0, 2\pi)$ are $\frac{\pi}{3}$ and $\frac{4\pi}{3}$.
If $\tan x+1=0$, then $\tan x=-1$, and the only solutions in $[0, 2\pi)$ are $\frac{3\pi}{4}$ and $\frac{7\pi}{4}$.
