Arccos 2x + arccos x = TT/2
How to solve ​find x ?

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

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