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 $sin^-1x=theta$

$=>sintheta=x$

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

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

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