For x ∈ ( - 1, 1 ] , the number of solutions of the equation ^ - 1...

Given that $\sin^{- 1}x = 2\tan^{- 1}x$

$⇒ \sin^{- 1}x = \sin^{- 1}\left(\dfrac{2x}{1 + {x}^{2}}\right)$

$⇒ x = \left(\dfrac{2x}{1 + {x}^{2}}\right)$

$⇒ x + {x}^{3} = 2x$

$⇒ {x}^{3} - x = 0$

$⇒ x ( {x}^{2} - 1 ) = 0$

$⇒ x = 0, - 1, 1$.

But $x \in ( - 1, 1 ]$, so the number of roots are $2$, $x = 0 \text{ and } 1$