Question

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

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

Answer 1

The correct option is A

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

Explanation for the correct option:

To find the value of $x$:

The given equation is $1+\sin x+{\sin }^{2}x+...\mathrm{\infty }=4+2\sqrt{3}$, $0<x<\pi$ and $x\ne \frac{\pi }{2}$,

$1+\sin x+{\sin }^{2}x+...\mathrm{\infty }$ is geometric progression with first term, $a=1$ and common ratio, $r=\sin x$ .

Sum of geometric progression upto $\mathrm{\infty }=\frac{a}{1-r}$, where $a$ is the first term and $r$ is the common ratio of geometric progression.

$\begin{array}{rl}1+\sin x+{\sin }^{2}x+...\mathrm{\infty }& =\frac{1}{1-\sin x}\\ 4+2\sqrt{3}& =\frac{1}{1-\sin x}\\ 1-\sin x& =\frac{1}{4+2\sqrt{3}}\\ 1-\sin x& =\frac{4-2\sqrt{3}}{(4+2\sqrt{3})(4-2\sqrt{3})}\\ & =\frac{4-2\sqrt{3}}{16-12}\\ & =\frac{4-2\sqrt{3}}{4}\\ & =1-\frac{\sqrt{3}}{2}\\ \sin x& =1-1+\frac{\sqrt{3}}{2}\\ \sin x& =\frac{\sqrt{3}}{2}\\ x& =\frac{\pi }{3},\frac{2\pi }{3}\end{array}$

Hence option(A): $\frac{\pi }{3},\frac{2\pi }{3}$ is the correct option.