Dear Vaibhav
Sin^-1(1-x) = (pi/2) + 2 sin^-1(x)
(1-x) = sin[pi/2 +(2sin^-1(x)]
1-x = cos (2 sin^-1(x))
On changing sin^-1 x to cos ^-1 x, and further solving
1-x = 1-2 x²
2 x² -x = 0
x = 0 and x = ½
But for x = 1/2 , LHS is not equal to RHS.
Hence
x = 0 is the solution.
Regards
Arun (askIITians forum expert)