Explanation

(c)
$\cos{2\theta = \sin\theta}$
$\Rightarrow \ 1 - 2\sin^{2}\theta = \sin\theta$
$\Rightarrow \ 2\sin^{2}\theta + \sin{\theta - 1 = 0}$
$\Rightarrow \ 2\sin^{2}{\theta + 2\sin\theta - \sin{\theta - 1 = 0}}$
$\Rightarrow \ 2\sin{\theta\left( \sin{\theta + 1} \right) - \left( \sin{\theta + 1} \right) = 0}$
$\Rightarrow \ \ \left( \sin\theta + 1 \right)\left( 2\sin{\theta - 1} \right) = 0$
$\Rightarrow \ \sin{\theta = - 1,\ \ \sin{\theta = \frac{1}{2}}}$
$\Rightarrow \ \sin{\theta = \sin{\frac{3\pi}{2},\sin{\theta = \sin\frac{\pi}{6}}}}$
$\Rightarrow \ \ \theta = n\pi + ( - 1)^{n}\frac{3\pi}{2},$
$\Rightarrow \ \ \ \theta = m\pi + ( - 1)^{m}\frac{\pi}{6}$
For $\theta \in (0,\ 2\pi)$
$\theta = \frac{3\pi}{2},\frac{\pi}{6},\frac{5\pi}{6}$
Hence number of solutions$= 3$