Question 518554
<pre>
cos<sup>-1</sup>(x) + cos<sup>-1</sup>(2x) = 60°

Let {{{alpha}}} = cos<sup>-1</sup>(x)
Let {{{beta}}} = cos<sup>-1</sup>(2x)

Then cos({{{alpha}}}) = x
and cos({{{beta}}}) = 2x

So the equation

cos<sup>-1</sup>(x) + cos<sup>-1</sup>(2x) = 60°

becomes 

{{{alpha}}} + {{{beta}}} = 60°

Take the cosine of both sides:

cos({{{alpha}}}+{{{beta}}}) = cos(60°)

Using an identity for cos({{{alpha}}}+{{{beta}}}),

cos({{{alpha}}})cos({{{beta}}}) - sin({{{alpha}}})sin({{{beta}}}) = {{{1/2}}}

Then since cos({{{alpha}}}) = x and cos({{{beta}}}) = 2x

x(2x) - sin({{{alpha}}})sin({{{beta}}}) = {{{1/2}}}

2x² - sin({{{alpha}}})sin({{{beta}}}) = {{{1/2}}}

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

sin²{{{theta}}} + cos²{{{theta}}} = 1

sin²{{{theta}}} = 1 - cos²{{{theta}}} 
          ___________ 
sin{{{theta}}} = ±<font face = "symbol">Ö</font>1 - cos²{{{theta}}}

The sign of the sine depends on what quadrants {{{alpha}}} and {{{beta}}} are in.  We consider all cases
           ___________     ______
sin(a) = ±<font face = "symbol">Ö</font>1 - cos²(a) = ±<font face = "symbol">Ö</font>1 - x²
           ___________     __________    _________ 
sin(b) = ±<font face = "symbol">Ö</font>1 - cos²(b} = ±<font face = "symbol">Ö</font>1 - (2x)² = ±<font face = "symbol">Ö</font>1 - 4x²

Substituting in

2x² - sin({{{alpha}}})sin({{{beta}}}) = {{{1/2}}}
       ______  _______
2x² ± <font face = "symbol">Ö</font>1 - x²·<font face = "symbol">Ö</font>1 - 4x² = {{{1/2}}}

       _________________
2x² ± <font face = "symbol">Ö</font>(1 - x²)(1 - 4x²) = {{{1/2}}}

Multiply both sides by 2 to clear of the fraction:

        _________________
4x² ± 2<font face = "symbol">Ö</font>(1 - x²)(1 - 4x²) = 1

Isolate the radical term:

         ___________________
      ±2<font face = "symbol">Ö</font>(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 = {{{(-3)/(-12)}}}

                        x² = {{{1/4}}}
                          
                         x = ±{{{1/2}}}

We must check for extraneous solutions:

Checking {{{1/2}}}:

cos<sup>-1</sup>(x) + cos<sup>-1</sup>(2x) = 60°

cos<sup>-1</sup>({{{1/2}}}) + cos<sup>-1</sup>(2·{{{1/2}}}) = 60°

60° + cos<sup>-1</sup>(1) = 60°

60° + 0° = 60°
   
60° = 60°

That checks, so {{{1/2}}} is a solution.

Checking {{{-1/2}}}:

cos<sup>-1</sup>(x) + cos<sup>-1</sup>(2x) = 60°

cos<sup>-1</sup>({{{-1/2}}}) + cos<sup>-1</sup>(2·{{{-1/2}}}) = 60°

120° + cos<sup>-1</sup>(-1) = 60°

120° + 180° = 60°
   
300° = 60°

That's false, so the only solution is 

x = {{{1/2}}}

Edwin</pre>