Let $\mathrm{A} = \left\{0 \in \left(-\frac{\mathrm{\pi }}{2}, \mathrm{\pi }\right): \frac{3+2\mathrm{isin\theta }}{1 -2\mathrm{isin\theta }} \mathrm{is} \mathrm{purely} \mathrm{imaginary}\right\}$  Then the sum of the element A is

$$\begin{gathered} \frac{3+2\mathrm{isin} \mathrm{\theta }}{1-2\mathrm{isin} \mathrm{\theta }} =\frac{(3+2\mathrm{isin} \mathrm{\theta })(1+2\mathrm{isin} \mathrm{\theta })}{(1-2\sin \mathrm{\theta })\left(1+2\mathrm{isin} \mathrm{\theta }\right)} \\ =\frac{\left(3-4{\sin }^2 \mathrm{\theta }\right)}{1+4{\sin }^2 \mathrm{\theta }}+\frac{\mathrm{i}(8\sin \mathrm{\theta })}{1+4\sin \mathrm{\theta }}\text{ is purely Imaginary.} \\ \therefore \text{ real part }=0\mathrm{ } \therefore 3-4{\sin }^2 \mathrm{\theta }=0 \\ \begin{array}{rr} {\sin }^2 \mathrm{\theta }& \mathrm{ }=\frac{3}{4}\Rightarrow \sin \mathrm{\theta }=\pm \frac{\sqrt{3}}{2},\mathrm{ }\mathrm{\theta }\in \left(-\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right)\end{array} \\ \begin{array}{rr}\mathrm{\theta }& \mathrm{ }=-\frac{\mathrm{\pi }}{3},\frac{\mathrm{\pi }}{3},\frac{2\mathrm{\pi }}{3}\Rightarrow \mathrm{set} \mathrm{A}=\left\{-\frac{\mathrm{\pi }}{3},\frac{\mathrm{\pi }}{3},\frac{2\mathrm{\pi }}{3}\right\}\end{array} \end{gathered}$$

Sum of elements of $\mathrm{A}=\frac{-\mathrm{\pi }}{3}+\frac{\mathrm{\pi }}{3}+\frac{2\mathrm{\pi }}{3}=\frac{2\mathrm{\pi }}{3}$