Explanation

(b)
Given, $2\sin^{2}{\frac{\theta}{2} = 2\sin{\frac{\theta}{2}.\cos{\frac{\theta}{2}.\sin\frac{\theta}{2}}}}$
$\Rightarrow \ 2\sin^{2}{\frac{\theta}{2}\left\lbrack 1 - \cos\frac{\theta}{2} \right\rbrack = 0}$
$\Rightarrow \ \sin{\frac{\theta}{2} = 0}$ or $2\sin^{2}{\frac{\theta}{4} = 0}$
$\Rightarrow \ \frac{\theta}{2} = k\pi$ or $\frac{\theta}{4} = k\pi$
Hence, $\theta = 2k\pi$ or $\theta = 4k\pi,\ \ \ k \in I$