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Answer 1 · 2017-04-03T18:52:20

Removed all content, because I liked dk_ch's answer much better than mine.

Removed all content, because I liked dk_ch's answer much better than mine.

Answer 2 · 2017-04-04T04:25:17

Given: $2sin^-1(x) + sin^-1(2x) = pi/2$

$=>2sin^-1(x) =pi/2 -sin^-1(2x)$

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

$=>1-2sin^2(sin^-1(x)) =sinsin^-1(2x) )$

$=>1-2x^2 =2x$

$=>2x^2 +2x-1=0$

$=>x=(-2pmsqrt(2^2-4*2*(-1)))/(2*2)$

$=>x=(-2pmsqrt12)/(2*2)$

$=>x=(-2pm2sqrt3)/(2*2)$

$=>x=(-1pmsqrt3)/2$

But $sin^-1(x)$ is feasible only when $-1<=x<=+1$
Again $x=(-1-sqrt3)/2<-1$ So this solution should be neglected and only solution is

$=>x=(-1+sqrt3)/2$

Answer 3 · 2017-04-04T04:41:59

$x=-1/2+sqrt3/2.$

Suppose that, $arcsinx=alpha, and, arcsin(2x)=beta.$

$:. sinalpha=x, and, sinbeta=2x.....(1).$

With these substns., the given eqn. becomes,

$2alpha+beta=pi/2, rArr beta=pi/2-2alpha.$

$rArr sinbeta=sin(pi/2-2alpha)=cos2alpha, i.e..$

$sinbeta=1-2sin^2alpha.$

$:. 2x=1-2x^2...........[because, (1).]$

$:. 2x^2+2x-1=0$

Using Quadr. Formula, we have, $x={-2+-sqrt(2^2-4(2)(-1))}/(2(2)),$

$:. x=(-2+-sqrt12)/4=(-2+-2sqrt3)/4=-1/2+-sqrt3/2.$

Among these, $x=-1/2-sqrt3/2$ does not satisfy the given eqn.

$:. x=-1/2+sqrt3/2," is the Soln.,"$ as readily derived by

Respected Douglas K. Sir!

Enjoy Maths.!