Answer
$\theta =-\dfrac {\pi }{2}+2\pi k$
Work Step by Step
$\sin ^{2}\theta =2\sin \theta +3\Rightarrow \sin ^{2}\theta -2\sin \theta -3=0$
$\sin \theta =\dfrac {2\pm \sqrt {2^{2}+4\times 3}}{2}=\dfrac {2\pm 4}{2}=3;-1$
$sin \theta$ can not be greater than 1 so
$\sin \theta =-1\Rightarrow \theta =\sin ^{-1}\left( -1\right) =-\dfrac {\pi }{2}+2\pi k$
