So from arccos(x) + arccos(2x) = 60º take the cosine from both sides
cos(arccos(x)+arccos(2x)) = 1/2
Expand the cosine sum, and use cos(arccos(theta)) = theta to rewrite it
x*2x - sin(arccos(x))sin(arccos(2x)) = 1/2
Use the pythagorean identity sin^2(x) = 1 - cos^2(x) to rewrite it. Knowing that the sine is always positive on the arccosine range.
x*2x - sqrt(1-x^2)*sqrt(1-4x^2)=1/2
2x^2 - sqrt((1-x^2)(1-4x^2)) = 1/2
Isolate the root, then square both sides
-sqrt((1-x^2)(1-4x^2)) = 1/2 - 2x^2
(1-x^2)(1-4x^2) = 1/4 - 2x^2 + 4x^4
Simplify,
1 - x^2 -4x^2 + 4x^4 = 1/4 - 2x^2 + 4x^4
1 -5x^2 = 1/4 -2x^2
1 - 1/4 = 5x^2 -2x^2
3/4 = 3x^2
1/4 = x^2
Taking the square root
x = +- 1/2
