Chapter 7 - Section 7.4 - Basic Trigonometric Equations - 7.4 Exercises - Page 569: 50

$\theta =\dfrac {\pi }{6}+\pi k$ $\theta =-\dfrac {\pi }{6}+\pi k$

$3\tan ^{3}\theta =\tan \theta \Rightarrow 3\tan ^{2}\theta =1\Rightarrow \tan \theta =\pm \dfrac {1}{\sqrt {3}}$ $\theta =\tan ^{-1}\left( \dfrac {1}{\sqrt {3}}\right) =\dfrac {\pi }{6}+\pi k$ $\theta =\tan ^{-1}\left( -\dfrac {1}{\sqrt {3}}\right) =-\dfrac {\pi }{6}+\pi k$