Which of the following is the solution set of the equation sin^-1x =…

For the existence of the given equation, we must have

-1 ≤ x ≤ 1 and -1 ≤ 3x -1 ≤ 1 ⇒ 0 ≤ x ≤ $\frac{2}{3}$

Now,

$sin^{-1}x = cos^{-1}x +sin^{-1}(3x-1)$

$⇒sin^{-1}x - cos^{-1}x =sin^{-1}(3x-1)$

$⇒ 2sin^{-1}x-\frac{\pi}{2}=sin^{-1}(3x-1)$

$⇒ sin\left(2sin^{-1}x-\frac{\pi}{2}\right) = sin \left(sin^{-1}(3x-1)\right)$

$⇒ -cos (2sin^{-1}x) = 3x -1 $

$⇒ - \begin{Bmatrix}1- 2 sin^2 (sin^{-1}x)\end{Bmatrix}= 3x - 1$

$⇒ -(1 -2x^2) = 3x - 1$

$⇒ 2x^2 -3x = 0 ⇒x = 0, \frac{3}{2} ⇒ x = 0 $        [∵ 0 ≤ x ≤ 3/2]