Question

Question #28bc1

Answer 1

`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`