If $1 + \mathrm{sin\theta }+{\sin }^2\mathrm{\theta }+\dots \infty =4+2\sqrt{3}, 0<\mathrm{\theta }<\mathrm{\pi }$ and $\mathrm{\theta }\ne \frac{\mathrm{\pi }}{2},$ then $\mathrm{\theta }$

$$\begin{gathered} 1+\mathrm{sin\theta }+{\sin }^2\mathrm{\theta }+......=4+2\sqrt{3} \\ \frac{1}{1-\mathrm{sin\theta }}=4+2\sqrt{3} \\ 1-\mathrm{sin\theta }=\frac{1}{2\left(2+\sqrt{3}\right)} \frac{2-\sqrt{3}}{2-\sqrt{3}} \\ 1-\mathrm{sin\theta }=\frac{2-\sqrt{3}}{2} \\ 1-\mathrm{sin\theta }=1-\frac{\sqrt{3}}{2} \\ \mathrm{sin\theta }=\frac{\sqrt{3}}{2} \\ \mathrm{\theta }=\frac{\mathrm{\pi }}{3}, \frac{2\mathrm{\pi }}{3} \end{gathered}$$