Question

If $1+\sin x+\sin^2 x+.......$ upto $\infty=4+2\sqrt{3},0 < x < \pi $ and $x \neq \frac {\pi}{2}$ 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, $1+\sin x+\sin ^{2} x+\ldots \infty=4+2 \sqrt{3}$
$\Rightarrow \frac{1}{1-\sin x}=4+2 \sqrt{3}$
$\Rightarrow 1-\sin x=\frac{1}{4+2 \sqrt{3}} \times \frac{4-2 \sqrt{3}}{4-2 \sqrt{3}}$
$\Rightarrow 1-\sin x=\frac{4-2 \sqrt{3}}{4}$
$\Rightarrow \sin x=\frac{2 \sqrt{3}}{4}=\frac{\sqrt{3}}{2}$
$\Rightarrow x=\frac{\pi}{3}, \frac{2 \pi}{3}$