1. We need to solve $\sin^{-1}x = 2\tan^{-1}x, \qquad x\in(-1,1].$

Let $\theta = \tan^{-1}x.$ Then $x = \tan\theta,$ and since $x\in(-1,1]$, we have $\theta \in \left(-\frac{\pi}{4},\frac{\pi}{4}\right].$

The equation becomes $\sin^{-1}x = 2\theta.$ Now take sine on both sides: $x = \sin(2\theta).$ But $x=\tan\theta$, so $\tan\theta = \sin 2\theta = 2\sin\theta\cos\theta.$

2. Write $\tan\theta$ in terms of sine and cosine: $\frac{\sin\theta}{\cos\theta} = 2\sin\theta\cos\theta.$ Multiply by $\cos\theta$: $\sin\theta = 2\sin\theta\cos^2\theta.$ So $\sin\theta\,(1-2\cos^2\theta)=0.$

Thus either

• $\sin\theta=0$, or
• $1-2\cos^2\theta=0$.

3. Case 1: $\sin\theta=0$. Then $\theta=0 \implies x=\tan 0=0.$ This is valid.

4. Case 2: $1-2\cos^2\theta=0$. Then $\cos^2\theta=\frac12.$ So $\theta = \pm \frac{\pi}{4} + k\pi.$ But $\theta\in\left(-\frac{\pi}{4},\frac{\pi}{4}\right]$, hence only $\theta=\frac{\pi}{4}$ is allowed. (Note: $-\frac{\pi}{4}$ is excluded because $x\in(-1,1]$ excludes $x=-1$.) Thus $x=\tan\frac{\pi}{4}=1.$ This is valid.

5. Check both solutions in the original equation:

• For $x=0$: $\sin^{-1}0=0, \qquad 2\tan^{-1}0=0.$ Valid.

• For $x=1$: $\sin^{-1}1=\frac{\pi}{2}, \qquad 2\tan^{-1}1=2\cdot\frac{\pi}{4}=\frac{\pi}{2}.$ Valid.

Hence the solutions are $x=0,\;1.$ Therefore, the number of solutions is $\boxed{2}.$