Here,
#Arcsin(1-x)-2arcsinx=pi/2#
#=>sin^-1(1-x)=pi/2+2sin^-1x#
#=>sin(sin^-1(1-x))=sin(pi/2+2sin^-1x)#
#=>1-x=cos(2sin^-1x)...to[as, sin(pi/2+alpha)=cosalpha]#
Let, #sin^-1x=theta=>x=sintheta#
so,
#1-sintheta=cos(2theta)#
#=>1-sintheta=1-2sin^2theta#
#=>2sin^2theta-sintheta==0#
#=>sintheta(2sintheta-1)=0#
#=>sintheta=0 or 2sintheta-1=0#
#=>x=0 or x=1/2#
CHECK:
#(1) If,x=0,then#
#LHS=arcsin(1-0)-2arcsin(0)=pi/2-2(0)=pi/2=RHS#
i.e. #x=0 # is the solution.
#(2)If, x=1/2,then#
#LHS=arcsin(1-1/2)-2arcsin(1/2)#
#=>LHS=arcsin(1/2)-2(pi/6)=pi/6-pi/3=-pi/6!=pi/2#
#=>LHS!=RHS#
i.e. #x!=1/2#
Hence, #x=0# is the solution