Explanation

(d)
Given, $1 + \sin{\theta + \sin^{2}{\theta + \ldots\infty = 4 + 2\sqrt{3}}}$
$\Rightarrow \ \ \frac{1}{1 - \sin\theta} = 4 + 2\sqrt{3}\ \ \ \ \lbrack\because 0 < \sin{\theta < 1}\rbrack$
$\Rightarrow \ \ 1 - \sin{\theta = \frac{4 - 2\sqrt{3}}{16 - 12} = 1 - \frac{\sqrt{3}}{2}}$
$\Rightarrow \ \sin{\theta = \frac{\sqrt{3}}{2}}$
$\Rightarrow \ \ \theta = \frac{\pi}{3}$ or $\frac{2\pi}{3}$