Answer
$\frac{\pi}{6}+k\pi$, $\frac{\pi}{4}+k\pi$, $\frac{5\pi}{6}+k\pi$
Work Step by Step
$3\tan^3 \theta-3\tan^2 \theta-\tan \theta+1=0$
$3\tan^2\theta(\tan \theta-1)-(\tan \theta-1)=0$
$(3\tan^2 \theta-1)(\tan \theta-1)=0$
$(\sqrt{3}\tan \theta+1)(\sqrt{3}\tan \theta-1)(\tan\theta-1)=0$
If $\sqrt{3}\tan \theta+1=0$, then $\sqrt{3}\tan \theta=-1$, and $\tan \theta=-\frac{\sqrt{3}}{3}$. The solutions are $\frac{5\pi}{6}+k\pi$.
If $\sqrt{3}\tan \theta-1=0$, then $\sqrt{3}\tan \theta=1$, and $\tan \theta=\frac{\sqrt{3}}{3}$. The solutions are $\frac{\pi}{6}+k\pi$.
If $\tan \theta-1=0$, then $\tan \theta=1$. The solutions are $\frac{\pi}{4}+k\pi$.
