You need to complete the square to the left side, such that:

`2cos^2 a - 3cos a = -1`

`2cos^2 a - 3cos a + (3/(2sqrt2))^2 = (3/(2sqrt2))^2 - 1`

`(sqrt2*cos a - 3/(2sqrt2))^2 = 9/8 - 1`

`(sqrt2*cos a - 3/(2sqrt2))^2 = 1/8 => sqrt2*cos a - 3/(2sqrt2)2 = +-sqrt(1/8) => sqrt2*cos a - 3/(2sqrt2) = +-1/(2sqrt2)`

`sqrt2*cos a = 3/(2sqrt2) +-1/(2sqrt2)`

`cos a = (3+-1)/4 => cos a = 1; cos a = 1/2`

`cos a = 1 => a = +-cos^(-1)0 + 2npi => a = +-pi/2 + 2npi`

`cos a = 1/2 => a = +-cos^(-1)(1/2) + 2npi => a = +-pi/3 + 2npi`

**Hence, evaluating the solutions to the given equation, yields `a = +-pi/2 + 2npi` and **`a = +-pi/3 + 2npi.`

The first step is to move all terms to the left side:

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

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

We'll note cos a = 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 a = t1.

cos a= 1

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

cos a = y

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

In our case, y = 1:

a = +/-arccos 1 + 2k*pi

a = 0 + 2k*pi

a = 2k*pi

Now, we'll put cos a = t2

cos a = 1/2

a = +/- arccos 1/2 + 2k*pi

a = +/- (pi/3) + 2k*pi

a = 2k*pi + pi/3

a = 2k*pi - pi/3

The solutions of the equation are:{ 2k*pi}U{2k*pi - pi/3}U{2k*pi + pi/3}