We're given the equation $\sin^{-1}x = 2\tan^{-1}x$ for $x$ in the interval $(-1, 1]$. We want to find the number of solutions.

Step 1: Apply the sine and tangent functions to both sides :

We can rewrite the equation by applying the sine function to both sides :

$$\sin(\sin^{-1}x) = \sin(2\tan^{-1}x).$$

This simplifies to:

$$x = \sin(2\tan^{-1}x).$$

Step 2: Use the double-angle identity for sine :

Recall that $\sin(2y) = 2\sin(y)\cos(y)$. Applying this identity to the right-hand side gives :

$$x = 2\sin(\tan^{-1}x)\cos(\tan^{-1}x).$$

Step 3: Use the identities for sine and cosine of an inverse tangent :

Recall that $\sin(\tan^{-1}x) = \frac{x}{\sqrt{1 + x^2}}$ and $\cos(\tan^{-1}x) = \frac{1}{\sqrt{1 + x^2}}$. Substituting these into the equation gives :

$$x = 2 \cdot \frac{x}{\sqrt{1 + x^2}} \cdot \frac{1}{\sqrt{1 + x^2}}.$$

This simplifies to :

$$x = \frac{2x}{1 + x^2}.$$

Step 4: Solve for $x$ :

We have :

$$x = \frac{2x}{1 + x^2}.$$

Cross-multiplying gives :

$$x(1 + x^2) = 2x.$$

This simplifies to :

$$x^3 + x - 2x = 0.$$

Rearranging terms gives :

$$x^3 - x = 0.$$

This factors to:

$$x(x^2 - 1) = 0.$$

Setting each factor equal to zero gives the solutions $x = 0$, $x = -1$, and $x = 1$.

However, we are given that $x \in (-1, 1]$. Therefore, the only solutions in this interval are $x = 0$ and $x = 1$.

So there are 2 solutions to the equation $\sin ^{-1} x=2 \tan ^{-1} x$ in the interval $x \in(-1,1]$.