Question

If $1+\sin x+{\sin }^{2}x+si{n}^{3}x+...\infty$is equal to $4+2\sqrt{3},0\le x<\pi$ then x is equal to

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$ or $\frac{\pi }{6}$
• D. $\frac{\pi }{3}$ or $\frac{2\pi }{3}$

Answer 1

The correct option is D $\frac{\pi }{3}$ or $\frac{2\pi }{3}$

$1+\sin x+{\sin }^{2}x+{\sin }^{3}x+.............+\infty =4+2\sqrt{3}$

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

$\Rightarrow 1-sinx=\frac{1}{4+2\sqrt{3}}.\frac{4-2\sqrt{3}}{4-2\sqrt{3}}=1-\frac{\sqrt{3}}{2}$

$\Rightarrow sinx=\frac{\sqrt{3}}{2}$

Hence solution in the given interval is,
$x=\frac{\pi }{3}or\frac{2\pi }{3}$
Hence, option 'D' is correct.