Question

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

• A. $\frac{\pi }{3},\frac{5\pi }{6}$
• B. $\frac{2\pi }{2},\frac{\pi }{6}$
• C. $\frac{\pi }{3},\frac{2\pi }{3}$
• D. $\frac{\pi }{6},\frac{\pi }{3}$

Answer 1

The correct option is C $\frac{\pi }{3},\frac{2\pi }{3}$

Given, $1+\sin x+{\sin }^{2}x+....\infty =4+2\sqrt{3}$
$\Rightarrow \frac{1}{1-\sin x}=4+2\sqrt{3}$
$\Rightarrow 1-\sin x=\frac{1}{4+2\sqrt{3}}=\frac{4-2\sqrt{3}}{4}$
$\Rightarrow \sin x=\frac{\sqrt{3}}{2}$
$\therefore x=\frac{\pi }{3},\frac{2\pi }{3}[\because x\in (0,\pi \left)\right]$