Explanation

(c)
We have, $\left( \frac{3 + 2i\sin\theta}{1 - 2i\sin\theta} \right) = \frac{(3 + 2i\sin{\theta)}}{(1 - 2i\sin\theta)}\frac{(1 + 2i\sin{\theta)}}{(1 + 2i\sin{\theta)}}$
$= \left( \frac{3 - 4\sin^{2}\theta}{1 + 4\sin^{2}\theta} \right) + i\left( \frac{8\sin\theta}{1 + 4\sin^{2}\theta} \right)$
Since, it is real therefore Im ($\mathcal{z}$) should be zero
$\Rightarrow \ \ \frac{8\sin\theta}{1 + 4\sin^{2}\theta} = 0\ \Rightarrow \ \sin{\theta = 0}$
$\therefore\ \ \theta = n\pi,\ where\ \ n = \ 0,\ 1,\ 2,\ 3,\ldots$