To solve the equation cos x - cos 3x = cos 2x - cos 4x, we can use the trigonometric identity:

cos a - cos b = -2sin((a+b)/2)sin((a-b)/2)

Substitute a = x and b = 3x:

cos x - cos 3x = -2sin((x+3x)/2)sin((x-3x)/2)
cos x - cos 3x = -2sin(2x/2)sin(-2x/2)
cos x - cos 3x = -2sin(x)sin(-x)
cos x - cos 3x = 2sin(x)sin(x)

Now let's simplify the right side using the double angle identity for sine:

cos x - cos 3x = 2sin^2(x)

Similarly, for cos 2x - cos 4x, we can follow the same steps:

cos a - cos b = -2sin((a+b)/2)sin((a-b)/2)

Substitute a = 2x and b = 4x:

cos 2x - cos 4x = -2sin((2x+4x)/2)sin((2x-4x)/2)
cos 2x - cos 4x = -2sin(3x)sin(-x)
cos 2x - cos 4x = 2sin(3x)sin(x)

Now, the equation looks like this:

2sin^2(x) = 2sin(3x)sin(x)

Divide both sides by 2:

sin^2(x) = sin(3x)sin(x)

Now use the trigonometric identity:

sin(a)sin(b) = (1/2)[cos(a-b) - cos(a+b)]

To get:

sin^2(x) = (1/2)[cos(3x-x) - cos(3x+x)]
sin^2(x) = (1/2)[cos(2x) - cos(4x)]

Therefore, cos x - cos 3x = cos 2x - cos 4x is true.