Question

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

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

$x-\frac{x^2}{2}+\frac{x^3}{4}-\frac{x^4}{8}+\cdots$ represents an infinite GP series with common ratio, $r=\frac{-x}{2}$.
$|r|=\left|\frac{x}{2}\right|<1$

Therefore, the equation can be simplified as
${\sin }^{-1}\left(\frac{x}{1-(-\frac{x}{2})}\right)=\frac{\pi }{6}\Rightarrow \frac{2x}{2+x}=\frac{1}{2}\Rightarrow x=\frac{2}{3}$