The solution set of the equation cos^-1 x - sin^-1x = sin^-1 (1-x), is

The correct answer is option 2: [0, 1/2]

We observe that the LHS of the given equation is defined for -1 ≤ x ≤ 1, whereas RHS   is defined for  -1 ≤ (1- x) ≤ 1 i.e 0 ≤ x ≤ 2. 

$cos^{-1} x - sin^{-1}x = sin^{-1} (1-x)$

$⇒ \frac{\pi}{2} -2 sin^{-1} x = sin^{-1}(1- x)$

$⇒ \frac{\pi}{2}- sin^{-1} (1 - x) = 2 sin^{-1} x $

$⇒ cos^{-1} (1-x) = cos^{-1}(1-2x^2)$

$⇒ 1- x = 1- 2x^2 ⇒2x^2 -x = 0 ⇒ x = 0, \frac{1}{2}$