(c)
$2\sin^{2}{\theta - \cos{2\theta = 0\ \ \Rightarrow \ \sin^{2}{\theta = \frac{1}{4}\ \ \ \Rightarrow \ \sin{\theta = \pm \frac{1}{2}}}}}$ …(i)
Also, $2\cos^{2}{\theta = 3\sin{\theta\ \ \Rightarrow \ 2\sin^{2}{\theta + 3\sin{\theta - 2 = 0}}}}$
$\Rightarrow \ \ \sin{\theta = \frac{1}{2}}$ …(ii)
From Eqs. (i) and (ii), $\sin{\theta = \frac{1}{2}}$
Two solutions exist in $\lbrack 0,\ 2\pi\rbrack$