The solution set of the equation ${\sin }^{-1}x=2{\tan }^{-1}x$ is:

We know the identity:
$$2{\tan }^{-1}x={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$$
Substituting this into the given equation:
$${\sin }^{-1}x={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$$
$$x=\frac{2x}{1+x^2}$$
$$x(1+x^2)=2x$$
$$x+x^3-2x=0$$
$$x^3-x=0$$
$$x(x^2-1)=0$$
$$x(x-1)(x+1)=0$$
Solutions are $x=0,1,-1$.The solution set is $\{-1,0,1\}$.