Q. If $1 + \sin x + \sin^{2}x + ...upto\ \infty = 4 + 2\sqrt{3},0 < x < \pi\ and\ x \neq \frac{\pi}{2}$, then $x$ is equal to
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) $\frac{\pi}{6},\frac{\pi}{3}$
Explanation
(c)
Given, $1 + \sin x + \sin^{2}x + ...\infty = 4 + 2\sqrt{3}$
$\Rightarrow \ \frac{1}{1 - \sin x} = 4 + 2\sqrt{3}$
$\Rightarrow 1 - \sin x = \frac{1}{4 + 2\sqrt{3}} \times \frac{4 - 2\sqrt{3}}{4 - 2\sqrt{3}}$
$\Rightarrow 1 - \sin x = \frac{4 - 2\sqrt{3}}{4}$
$\Rightarrow \ \sin x = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$
$\Rightarrow \ x = \frac{\pi}{3},\frac{2\pi}{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.