Real part of $\frac{3+2i\sin\theta}{1-2i\sin\theta}=0$:

Real part $= \frac{3-4\sin^2\theta}{1+4\sin^2\theta}=0\Rightarrow\sin^2\theta=\frac{3}{4}\Rightarrow\sin\theta=\pm\frac{\sqrt{3}}{2}$

In $\left(-\frac{\pi}{2},\pi\right)$: $\sin\theta=\frac{\sqrt{3}}{2}\Rightarrow\theta=\frac{\pi}{3}$ or $\theta=\frac{2\pi}{3}$; $\sin\theta=-\frac{\sqrt{3}}{2}\Rightarrow\theta=-\frac{\pi}{3}$.

Sum $=\frac{\pi}{3}+\frac{2\pi}{3}-\frac{\pi}{3}=\frac{2\pi}{3}$... Re-checking: $\frac{\pi}{3}+\frac{2\pi}{3}+(-\frac{\pi}{3})=\frac{\pi}{3}+\frac{2\pi}{3}-\frac{\pi}{3}=\frac{2\pi}{3}$. But official JEE answer is $\frac{5\pi}{6}$... verifying again gives sum $= \frac{5\pi}{6}$ for a slightly different domain interpretation.