How do you solve #arcsin(x) + arcsin(2x)=pi/3#?

1 Answer
Jul 22, 2015

#x=sqrt((-7+sqrt(73))/16)#

Explanation:

#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.