Answer
$\theta =\dfrac {7\pi }{6}+2\pi k;\theta =\dfrac {11\pi }{6}+2\pi k$
$\theta =\dfrac {\pi }{2}+2\pi k, \theta =\dfrac {3 \pi }{2}+2\pi k, $
Work Step by Step
$\cos \theta \left( 2\sin \theta +1\right) =0$
$\cos \theta =0\Rightarrow \theta =\cos ^{-1}\left( 0\right) =\dfrac {\pi }{2}+2\pi k, \theta =\dfrac {3 \pi }{2}+2\pi k, $
$
2\sin \theta +1=0\Rightarrow \sin \theta =-\dfrac {1}{2}\Rightarrow \theta =\sin ^{-1}\left( -\dfrac {1}{2}\right) \Rightarrow \theta =\dfrac {7\pi }{6}+2\pi k;\theta =\dfrac {11\pi }{6}+2\pi k$
