arccos(2x) + arccos(x) = π/2
Apply cos to both sides, then expand the right side using the identity I mentioned in a comment:
cos(arccos(2x) + arccos(x)) = cos(π/2)
cos(arccos(2x)) cos(arccos(x)) - sin(arccos(2x)) sin(arccos(x)) = 0
Now think of arccos(2x) and arccos(x) as angles in two right triangles. Let θ = arccos(2x) and φ = arccos(x) (and let's also assume that both have measure between 0 and π/2). These are angles such that
cos(θ) = 2x
cos(φ) = x
In these triangles, you then have
sin(θ) = sin(arccos(2x)) = √(1 - 4x ²)
sin(φ) = sin(arccos(x)) = √(1 - x ²)
while
cos(θ) = cos(arccos(2x)) = 2x
cos(φ) = cos(arccos(x)) = x
So the original equation is transformed to
2x ² - √(1 - 4x ²) √(1 - x ²) = 0
Solve for x :
2x ² = √(1 - 4x ²) √(1 - x ²)
(2x ²)² = (√(1 - 4x ²) √(1 - x ²))²
4x ⁴ = (1 - 4x ²) (1 - x ²)
4x ⁴ = 1 - 5x ² + 4x ⁴
0 = 1 - 5x ²
x ² = 1/5
x = ± 1/√5
Now bearing in mind that we assume 0 < φ < π/2, we should have cos(φ) = x > 0, so we take the positive solution,
x = 1/√5
