Question

If $0<\sin x<1$ and $1+\sin x+(\sin x{)}^2+$......... upto infinity $=2$, then $x$ is equal to

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. none of these

Answer 1

Answer: (A)

$\frac{\pi }{6}$

The above forms a G.P.
The sum of this infinite G.P is $\frac{1}{1-{\sin }^{-1}\left(x\right)}=2$
$\Rightarrow \frac{1}{2}=1-{\sin }^{-1}\left(x\right)$

$\Rightarrow {\sin }^{-1}\left(x\right)=\frac{1}{2}$

$\Rightarrow x=\sin \left(\frac{1}{2}\right)$

$\Rightarrow \therefore x=\frac{\pi }{6}$