Question

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

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

Answer 1

Answer: (D)

Given, $1+\sin x+{\sin }^{2}x+\dots \infty =4+2\sqrt{3}$
$\Rightarrow \frac{1}{1-\sin x}=4+2\sqrt{3}$
$\Rightarrow 1-\sin x=\frac{1}{4+2\sqrt{3}}\times \frac{4-2\sqrt{3}}{4-2\sqrt{3}}$
$\Rightarrow 1-\sin x=\frac{4-2\sqrt{3}}{4}$
$\Rightarrow \sin x=\frac{2\sqrt{3}}{4}=\frac{\sqrt{3}}{2}$
$\Rightarrow x=\frac{\pi }{3},\frac{2\pi }{3}$