Question

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

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

Answer 1

Answer: (C)

$\frac{\pi }{3},\frac{2\pi }{3}$

Given, $1+\sin x+{\sin }^{2}x+....\infty =4+2\sqrt{3},0<x<\pi$

$\Rightarrow \frac{1}{1-\sin x}=4+2\sqrt{3}$ ..... [Using sum of infinite GP]

$\Rightarrow 1-\sin x=\frac{1}{4+2\sqrt{3}}\times \frac{4-2\sqrt{3}}{4-2\sqrt{3}}$

$=\frac{4-2\sqrt{3}}{16-12}=\frac{4-2\sqrt{3}}{4}$

$\Rightarrow \sin x=1-\frac{4-2\sqrt{3}}{4}=\frac{2\sqrt{3}}{4}$

$\Rightarrow \sin x=\frac{\sqrt{3}}{2}$

$\Rightarrow x=\frac{\pi }{3},\frac{2\pi }{3}$