If sin−1⁡2x1−x2−2sin−1⁡x=0, then x belongs to the interval

Let $\sin^{- 1} x = θ .$ Then, $x = \sin θ$ and $\sqrt{1 - x^{2}} = \cos θ$

Now,

$\sin^{- 1} (2 x \sqrt{1 - x^{2}})$

$= \sin^{- 1} (\sin 2 θ) = 2 θ,$ if $- \frac{π}{2} \leq 20 \leq \frac{π}{2}$

$= 2 \sin^{- 1} x,$ if $- \frac{π}{4} \leq θ \leq \frac{π}{4}$ i.e. if $- \frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}$

$∴ \sin^{- 1} (2 x \sqrt{1 - x^{2}}) - 2 \sin^{- 1} x = 0,$ if $- \frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}$

If $\sin^{- 1} (2 x \sqrt{1 - x^{2}}) - 2 \sin^{- 1} x = 0,$ then $x$ belongs to the interval