For $$x \in (-1, 1]$$, the number of solutions of the equation $$\sin^{-1}x = 2\tan^{-1}x$$ is equal to _____.

We need to find the number of solutions of $$\sin^{-1}x = 2\tan^{-1}x$$ for $$x \in (-1, 1]$$.

We start by noting that

Let $$\theta = \tan^{-1}x$$, so $$x = \tan\theta$$. Since $$x \in (-1, 1]$$, we have $$\theta \in (-\pi/4, \pi/4]$$.

The equation becomes: $$\sin^{-1}(\tan\theta) = 2\theta$$

Taking sine of both sides: $$\tan\theta = \sin(2\theta)$$

Next,

Using $$\sin(2\theta) = 2\sin\theta\cos\theta$$:

$$ \frac{\sin\theta}{\cos\theta} = 2\sin\theta\cos\theta $$

$$\sin\theta = 0$$, i.e., $$\theta = 0$$, giving $$x = 0$$.

Check: $$\sin^{-1}(0) = 0$$ and $$2\tan^{-1}(0) = 0$$. Valid solution.

$$\sin\theta \neq 0$$. Divide both sides by $$\sin\theta$$:

$$ \frac{1}{\cos\theta} = 2\cos\theta $$

$$ \cos^2\theta = \frac{1}{2} $$

$$ \cos\theta = \frac{1}{\sqrt{2}} \quad (\text{positive since } \theta \in (-\pi/4, \pi/4]) $$

$$ \theta = \pm\frac{\pi}{4}, \quad \text{giving } x = \pm 1 $$

From this,

For $$x = 1$$ (which is in the domain $$(-1, 1]$$):

$$\sin^{-1}(1) = \pi/2$$ and $$2\tan^{-1}(1) = 2(\pi/4) = \pi/2$$. Valid.

For $$x = -1$$: this is NOT in the domain $$(-1, 1]$$ (the domain is open at $$-1$$). So $$x = -1$$ is excluded.

The valid solutions are $$x = 0$$ and $$x = 1$$, giving 2 solutions.