Explanation

(c)
Let $z = \frac{3 + 2i\sin\theta}{1 - 2i\sin\theta}$
$\Rightarrow \ \ \ \ \ \ \ z = \frac{3 + 2i\sin\theta}{1 - 2i\sin\theta} \times \frac{(1 + 2i\sin\theta)}{1 + 2i\sin\theta}$
$= \frac{3 - 4\sin^{2}{\theta + 8i\sin\theta}}{1 + 4\sin^{2}\theta}$
For purely imaginary of $z$, put Re$(z) = 0$
$ie,\ \ \ \ \ \ \ \ \frac{3 - 4\sin^{2}\theta}{1 + 4\sin^{2}\theta} = 0$
$\Rightarrow \ \ \ \ \ \ \sin{\theta = \pm}\frac{\sqrt{3}}{2}\$
$\ \ \ \ \Rightarrow \ \ \ \ \theta = n\pi + ( - 1)^{n}\left( + \frac{\pi}{3} \right) = n\pi \pm \frac{\pi}{3}$