plz solve

sin^-1 (1-x) - 2 sin^-1x  = /2

sin^-1 (1-x) - 2 sin^-1x  =  sin^-1 (1-x)  +  cos^-1 (1-x) 

                                     (  sin^-1x  +  cos^-1x  = /2)

 - 2 sin^-1x  =  cos^-1 (1-x)

Now,let  sin^-1x = a,then  -2a = cos^-1 (1-x)

  or        cos2a =  1 - x               cos(-x) = cosx 

        1 - 2 sin²a = 1 - x            1 - 2x² = 1 - x   

                                                                         (  sin a = x )

    2x² - x = 0           x = 0 , 1/2

      ^{x = 1/2 is avoided, as it is not true.}

    ^{x = 0 is the only solution.}