#arcsin(x)+arcsin(2x)=pi/3#
Start by letting #alpha=arcsin(x)" "# and #" "beta=arcsin(2x)#
#color(black)alpha# and #color(black)beta# really just represent angles.
So that we have : #alpha+beta=pi/3#
#=>sin(alpha)=x#
#cos(alpha)=sqrt(1-sin^2(alpha))=sqrt(1-x^2)#
Similarly,
#sin(beta)=2x#
#cos(beta)=sqrt(1-sin^2(beta))=sqrt(1-(2x)^2)=sqrt(1-4x^2)#
#color(white)#
Next, consider
#alpha+beta=pi/3#
#=>cos(alpha+beta)=cos(pi/3)#
#=>cos(alpha)cos(beta)-sin(alpha)sin(beta)=1/2#
#=>sqrt(1-x^2)*sqrt(1-4x^2)-(x)*(2x)=1/2#
#=>sqrt(1-4x^2-x^2-4x^4)=2x^2+1/2#
#=>[sqrt(1-4x^2-x^2-4x^4)]^2=[2x^2+1/2]^2#
#=>1-5x^2-4x^4=4x^4+2x^2+1/4#
#=>8x^4+7x^2-3/4=0#
#=>32x^4+28x^2-3=0#
Now apply the quadratic formula in the variable #x^2#
#=>x^2=(-28+-sqrt(784+384))/64=(-28+-sqrt(1168))/64=(-28+-sqrt(16*73))/64=(-7+-sqrt(73))/16#
#=>x=+-sqrt((-7+-sqrt(73))/16)#
#color(white)#
Failed cases :
#color(red)((1)" .. "##x=+-sqrt((-7-sqrt(73))/16)#
is to be rejected because the solution is complex # inZZ#
#color(red)((2)" .. "##x=-sqrt((-7+sqrt(73))/16)#
is rejected because the solution is negative. Whereas #pi/3# is positive.
