Answer
$\frac{\pi}{6}, \frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6} $
Work Step by Step
Using trigonometric identities, we have
$tan^2\theta=\frac{1}{3} \longrightarrow tan\theta=\pm\frac{\sqrt 3}{3} \longrightarrow \theta=\frac{\pi}{6}, \pi-\frac{\pi}{6},\pi+\frac{\pi}{6},2\pi-\frac{\pi}{6} \longrightarrow \theta=\frac{\pi}{6}, \frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6} $