Arcsin(1-x)-2arcsinx=pi/2, then x is equal to ?

1 Answer
Apr 16, 2018

#x=0#

Explanation:

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