Question #28bc1

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Answer 1 · 2017-03-26T09:16:36

#2sin^-1x-3cos^-1x=pi/6#

Putting #color(blue)(cos^-1x=pi/2-sin^-1x)#

#=>2sin^-1x-3(pi/2-sin^-1x)=pi/6#

#=>2sin^-1x-(3pi)/2+3sin^-1x=pi/6#

#=>5sin^-1x=(3pi)/2+pi/6=(10pi)/6=(5pi)/3#

#=>5sin^-1x=(5pi)/3#

#=>sin^-1x=pi/3#

#=>x=sin(pi/3)=sqrt3/2#

please note

Let #sin^-1x=theta#

#=>sintheta=x#

#=>cos(pi/2-theta)=x#

#=>pi/2-theta=cos^-1x#

#=>pi/2-sin^-1x=cos^-1x#