The solutions of the equation sin^-1 x = 2 tan^-1x.

Clearly , LHS of the given equation is meaningful for x ∈ [-1, 1] and RHS is defined for all x ∈ R. So, the given equation exists for x ∈ [-1,1].

Now,

$sin^{-1}x = 2 tan^{-1}x $

$⇒sin^{-1}x = sin^{-1} \left(\frac{2x}{1+x^2}\right)$

$⇒ x^3 = x ⇒ x =0 , 1, -1.$