Answer
$\theta =\dfrac {\pi }{6}+\pi k;$
$\theta =\dfrac {5\pi }{6}+\pi k$
Work Step by Step
$3\tan ^{2}\theta -1=0\Rightarrow \tan \theta =\pm \dfrac {1}{\sqrt {3}}$
$\tan \theta =\dfrac {1}{\sqrt {3}}\Rightarrow \theta =\tan ^{-1}\left( \dfrac {1}{\sqrt {3}}\right) =\dfrac {\pi }{6}+\pi k$
$\tan \theta =-\dfrac {1}{\sqrt {3}}\Rightarrow \theta =\tan ^{-1}\left(- \dfrac {1}{\sqrt {3}}\right) =\dfrac {5 \pi }{6}+\pi k$
