You need to perform the following replacement of variable, such that:

`cos x = y `

Replacing the variable x yields:

`2y^2 + y - 1 = 0 => y^2 + y^2 + y - 1 = 0`

You need to form two groups of terms such that:

`(y^2 - 1) + (y^2 + y) = 0`

Converting the difference of squares `y^2 - 1` into a product yields:

`(y - 1)(y + 1) + y(y + 1) = 0`

Factoring out `(y + 1)` yields:

`(y + 1)(y - 1 + y) = 0 => {(y + 1 = 0),(2y - 1 = 0):} => (y = -1),(y = 1/2):}`

Replacing back `cos x` for y yields:

`cos x = -1 => x = +-cos^(-1)(-1) + 2n*pi`

`x = pi +- cos^(-1)1 + 2n*pi => x = pi +- 0 + 2n*pi => x = pi + 2n*pi`

`cos x = 1/2 => x = +-cos^(-1)(1/2) + 2n*pi => x = +-pi/3 + 2n*pi`

**Hence, evaluating the solutions to the given equation yields `x = pi + 2n*pi` and **`x = +-pi/3 + 2n*pi, n in Z.`

We'll re-write the equation changing the variable:

Let cos x = t => (cos x)^2 = t^2

We'll re-write the equation:

2t^2 + t - 1 = 0

Now, we'll apply the quadratic formula:

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

t1 = (-1+3)/4

t1 = 2/4

t1 = 1/2

t2 = (-1-3)/4

t2 = -1

Now, we'll find x from:

cos x = t1

cos x = 1/2

x = arccos (1/2) + 2kpi

x = pi/3 + 2kpi

cos x = t2

cos x = -1

x = arccos(-1) + 2kpi

x = pi + 2kpi

**The solutions of the equation are found in the following sets: {pi/3 + 2kpi}U{pi + 2kpi}.**