Question

If ${\sin }^{-1}(x-\frac{x^2}{2}+\frac{x^3}{4}-\frac{x^4}{8}+...)=\frac{\pi }{6}$, where $|x|<2$, then the value of $x$ is

• A. $\frac{2}{3}$
• B. $\frac{3}{2}$
• C. $-\frac{2}{3}$
• D. $-\frac{3}{2}$

Answer 1

The correct option is A $\frac{2}{3}$

We have,

${\sin }^{-1}(x-\frac{x^2}{2}+\frac{x^3}{4}-\frac{x^4}{8}+....)=\frac{\pi }{6}$

$\Rightarrow {\sin }^{-1}\left(\frac{x}{1-\left(\frac{-x}{2}\right)}\right)=\frac{\pi }{6}$ $[\because s_{\infty }=\frac{a}{1-r}]$

$\Rightarrow {\sin }^{-1}\left(\frac{2x}{2+x}\right)=\frac{\pi }{6}$

$\Rightarrow \frac{2x}{2+x}=\sin \frac{\pi }{6}\Rightarrow \frac{2x}{2+x}=\frac{1}{2}$

$\Rightarrow 4x=2+x\Rightarrow 3x=2\Rightarrow x=\frac{2}{3}$