You need to use the following trigonmetric identity, such that:

`cos x = cos y => x = +- cos^(-1)(cos y) + 2n*pi, n in Z`

`x = +-y + 2n*pi`

Hence, reasoning by analogy, yields:

`cos 2x=cos x => 2x = +-x + 2n*pi`

`2x = x + 2n*pi => 2x - x = 2n*pi => x = 2n*pi`

`2x =- x + 2n*pi => 2x + x = 2n*pi => 3x = 2n*pi => x = 2n*(pi/3)`

**Hence, evaluating the solutions to the given equation yields `x = 2n*pi` and **`x = 2n*(pi/3).`

We'll shift cos x to the left:

cos 2x - cos x = 0

Now, we'll apply the double angle identity:

cos 2x = cos (x+x) = cos x*cos x - sin x*sin x

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

We'll write (sin x)^2 = 1 - (cos x)^2 (fundamental formula of trigonometry).

cos 2x = (cos x)^2 - [1 - (cos x)^2]

We'll remove the brackets:

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

We'll combine like terms:

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

We'll re-write the equation, substituting cos 2x y the expression (1).

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

Now , we'll use substitution technique to solve the equation.

We'll note cos x = t and we'll re-write the equation in t:

2t^2 - t - 1 = 0

Since it is a quadratic, we'll apply the quadratic formula:

t1 = {-(-1) + sqrt[(-1)^2 + 4*2*1]}/2*2

t1 = [1+sqrt(1+8)]/4

t1 = (3+1)/4

t1 = 1

t2 = (1-3)/4

t2 = -1/2

Now, we'll put cos x = t1.

cos x = 1

Since it is an elementary equation, we'll apply the formula:

cos x = a

x = +/- arccos a + 2k*pi

x = arccos 1 + 2k*pi

x = 0

x = 2pi

cos x = t2

cos x = -1/2

x = pi - pi/3

x = 2pi/3

x = pi + pi/3

x = 4pi/3

The solutions for the equation are:{0 ; 2pi/3 ; 4pi/3 ; 2pi}.