Question

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

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

Answer 1

The correct option is C $\frac{\pi }{3},\frac{2\pi }{3}$

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

Using $a+ar+a{r}^2.......=\frac{a}{1-r}$

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

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

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

So, we get

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

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