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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.!