If sin ^{-1}left(x-frac{x^{2}}{2}+frac{x^{3}} | vedclass

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(A) Given, $\sin ^{-1}\left(x-\frac{x^{2}}{2}+\frac{x^{3}}{4}-\frac{x^{4}}{8}+\ldots\right)=\frac{\pi}{6}$.This is an infinite geometric series inside

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$\frac{2}{3}$

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(A) Given, $\sin ^{-1}\left(x-\frac{x^{2}}{2}+\frac{x^{3}}{4}-\frac{x^{4}}{8}+\ldots\right)=\frac{\pi}{6}$.This is an infinite geometric series inside the $\sin ^{-1}$ function with first term $a = x$ and common ratio $r = -\frac{x}{2}$.The sum of an infinite geometric series is given by $S_{\infty} = \frac{a}{1-r}$.Substituting the values, we get $S_{\infty} = \frac{x}{1 - (-\frac{x}{2})} = \frac{x}{1 + \frac{x}{2}} = \frac{2x}{2+x}$.So, $\sin ^{-1}\left(\frac{2x}{2+x}\right) = \frac{\pi}{6}$.Taking $\sin$ on both sides, we get $\frac{2x}{2+x} = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$.Cross-multiplying, we get $4x = 2 + x$.$3x = 2$, which gives $x = \frac{2}{3}$.

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

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