The first step is to move all terms to one side:

2(cos x)^2- 3cos x + 1 = 0

We'll solve the equation using substitution technique.

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

2t^2 - 3t + 1 = 0

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

t1 = {-(-3) + sqrt[(-3)^2 - 4*2*1]}/2*2

t1 = [3+sqrt(9-8)]/4

t1 = (3+1)/4

t1 = 1

t2 = (3-1)/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 = (-1)^k* arccos a + 2k*pi

In our case, a = 1:

x = (-1)^k* arccos 1 + 2k*pi

x = (-1)^k*(0) + 2k*pi

x = 2k*pi or x = 2pi + 2k*pi

**x = 0**

**x = 2pi(k+1)**

**x = 2pi**

Now, we'll put cos x = t2

cos x = 1/2

x = (-1)^k* arccos 1/2 + 2k*pi

x = (-1)^k* (pi/3) + 2k*pi

x = pi/3

x = pi - pi/3

x = 2pi/3

The solutions of the equation are:{ 0 ; pi/3 ; 2pi/3 ; 2pi }