Question

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

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

Answer 1

Answer: (A)

We have, $\sin ^{-1}\left(x-\frac{x^{2}}{2}+\frac{x^{3}}{4}-\frac{x^{4}}{8}+\ldots\right)=\frac{\pi}{6}$
$\Rightarrow \sin ^{-1}\left(\frac{x}{1-\left(\frac{-x}{2}\right)}\right)=\frac{\pi}{6}$ $\left[\because S_{\infty}=\frac{a}{1-r}\right]$
$\Rightarrow \sin ^{-1}\left(\frac{2 x}{2+x}\right)=\frac{\pi}{6}$
$\Rightarrow \frac{2 x}{2+x}=\sin \frac{\pi}{6}$
$\Rightarrow \frac{2 x}{2+x}=\frac{1}{2}$
$\Rightarrow 4 x=2+ x \Rightarrow 3 x=2$
$\Rightarrow x=\frac{2}{3}$