Answer
$\theta =\sin ^{-1}\left( 1\right) =\dfrac {\pi }{2}+2\pi k$
$\theta =\sin ^{-1}\left( \dfrac {-1}{2}\right) \Rightarrow \theta =-\dfrac {\pi }{6}+2\pi k$
Work Step by Step
$2\sin ^{2}\theta -\sin \theta -1=0 \Rightarrow \sin \theta =\dfrac {1\pm \sqrt {1-4\times 2\times \left( -1\right) }}{2\times 2}=\dfrac {1\pm 3}{4}=1;\dfrac {-1}{2} $
$\theta =\sin ^{-1}\left( 1\right) =\dfrac {\pi }{2}+2\pi k$
$\theta =\sin ^{-1}\left( \dfrac {-1}{2}\right) \Rightarrow \theta =-\dfrac {\pi }{6}+2\pi k$
