We have to find x for which cos 4x = cos 2x

we use the relation cos 2x = 2(cos x)^2 - 1

cos 4x = cos 2x

=> 2(cos 2x)^2 - 1= cos 2x

=> 2(cos 2x)^2 - cos 2x - 1 = 0

=> 2(cos 2x)^2 - 2cos 2x + cos 2x - 1 = 0

=> 2cos 2x( cos 2x - 1) + 1( cos 2x - 1) = 0

=> ( 2cos 2x + 1)( cos 2x - 1) = 0

( 2cos 2x + 1) = 0

=> cos 2x = -1/2

=> 2x = 120 degrees

=> x = 60 degrees

cos 2x - 1 = 0

=> cos 2x = 1

=> x = 0

**Therefore x = 0 + n*360 degrees and 60+ n*360 degrees**

To solve this equation, we'll move all the terms to the left side:

cos 4x - cos 2x = 0

Since the trigonometric functions are matching , we'll transform the difference into a product.

2 sin[(4x+2x)/2]*sin [(2x-4x)/2]=0

2 sin3x* sin (-x)=0

From this product of 2 factors, one or the other factor is zero.

sin3x=0

This is an elementary equation:

3x = (-1)^k*arcsin 0 + k*pi

3x = k*pi

x = k*pi/3, where k is an integer number.

We'll solve the second elementary equation:

sin (-x)=0

-sin (x)=0

sin x=0

x=(-1)^k arcsin 0 + k*pi

x=k*pi

**The set of solutions is : S={k*pi/3}U{k*pi}.**