cos-1(x) + cos-1(2x) = 60°

Let  = cos-1(x)
Let  = cos-1(2x)

Then cos() = x
and cos() = 2x

So the equation

cos-1(x) + cos-1(2x) = 60°

becomes 

 +  = 60°

Take the cosine of both sides:

cos(+) = cos(60°)

Using an identity for cos(+),

cos()cos() - sin()sin() = 

Then since cos() = x and cos() = 2x

x(2x) - sin()sin() = 

2x² - sin()sin() = 

Now we have to find the sines.  We use the identity:

sin² + cos² = 1

sin² = 1 - cos² 
          ___________ 
sin = ±Ö1 - cos²

The sign of the sine depends on what quadrants  and  are in.  We consider all cases
           ___________     ______
sin(a) = ±Ö1 - cos²(a) = ±Ö1 - x²
           ___________     __________    _________ 
sin(b) = ±Ö1 - cos²(b} = ±Ö1 - (2x)² = ±Ö1 - 4x²

Substituting in

2x² - sin()sin() = 
       ______  _______
2x² ± Ö1 - x²·Ö1 - 4x² = 

       _________________
2x² ± Ö(1 - x²)(1 - 4x²) = 

Multiply both sides by 2 to clear of the fraction:

        _________________
4x² ± 2Ö(1 - x²)(1 - 4x²) = 1

Isolate the radical term:

         ___________________
      ±2Ö(1 - x²)(1 - 4x²) = 1 - 4x²

Square both sides:

        4(1 - x²)(1 - 4x²) = (1 - 4x²)²

          4(1 - 5x² + 4x4) = 1 - 8x² + 16x4

           4 - 20x² + 16x4 = 1 - 8x² + 16x4

                     -12x² = -3        
                          
                        x² = 

                        x² = 
                          
                         x = ±

We must check for extraneous solutions:

Checking :

cos-1(x) + cos-1(2x) = 60°

cos-1() + cos-1(2·) = 60°

60° + cos-1(1) = 60°

60° + 0° = 60°
   
60° = 60°

That checks, so  is a solution.

Checking :

cos-1(x) + cos-1(2x) = 60°

cos-1() + cos-1(2·) = 60°

120° + cos-1(-1) = 60°

120° + 180° = 60°
   
300° = 60°

That's false, so the only solution is 

x = 

Edwin