Thank you for reporting, we will resolve it shortly
Q.
If $2\,sin^{-1}\,x = sin^{-1}\left(2x\sqrt{1-x^{2}}\right)$, then $x$ belongs to
Inverse Trigonometric Functions
Solution:
Since, $2\,sin^{-1}\,x = sin^{-1}\left(2x\sqrt{1-x^{2}}\right)$
Range of right hand side is $\left[-\frac{\pi}{2},\frac{\pi }{2}\right]$.
$\Rightarrow -\frac{\pi }{2} \le2\,sin^{-1}\,x \le\frac{\pi }{2}$
$\Rightarrow -\frac{\pi }{4} \le sin^{-1}\,x \le \frac{\pi }{4} $
$\Rightarrow x \in \left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right]$