Question

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

Answer 1

`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.