The number of solutions of the equation, $$\sin^{-1}x = 2\tan^{-1}x$$ (in principal values) is :

Let $$\tan^{-1} x = \theta$$. This implies that $$x = \tan \theta$$.

For principal values of the inverse tangent function, $$\theta$$ must be in the interval $$(-\frac{\pi}{2}, \frac{\pi}{2})$$.

$$\sin^{-1} x = 2\theta$$

$$x = \sin(2\theta)$$

Since $$x = \tan \theta$$ and $$x = \sin(2\theta)$$, $$\tan \theta = \sin(2\theta)$$

Using the identity $$\sin(2\theta) = \frac{2 \tan \theta}{1 + \tan^2 \theta}$$:

$$\tan \theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}$$

$$\tan \theta \left(1 - \frac{2}{1 + \tan^2 \theta}\right) = 0$$

Case 1: $$\tan \theta = 0$$

$$\implies \theta = 0$$

$$\implies x = \tan(0) = \mathbf{0}$$

Case 2: $$1 - \frac{2}{1 + \tan^2 \theta} = 0$$

$$\implies 1 = \frac{2}{1 + \tan^2 \theta}$$

$$\implies 1 + \tan^2 \theta = 2$$

$$\implies \tan^2 \theta = 1$$

$$\implies \tan \theta = \pm 1$$

$$\implies x = \mathbf{1}$$ or $$x = \mathbf{-1}$$

There are exactly 3 solutions: $$x \in \{ -1, 0, 1 \}$$.