Find to the nearest degree the solution set to ; cos² x = 2cos x +1

Domain is 0 < x < 360
Also, show work please

We treat the equation as a standard quadratic and solve it:
cos²x = 2cosx + 1
cos²x - 2cosx - 1 = 0
(cosx - 1)² - 2 = 0
(cosx - 1)² = 2
cosx - 1 = ±√2
cosx = 1 ± √2
-1 < cosx < 1, so cosx = 1 - √2
xᵣ = cosˉ¹(√2 - 1)
xᵣ = 65.53°
x = (180 - xᵣ)°, (180 + xᵣ)°
x = (180 - 65.53)°, (180 + 65.53)°
x = 114.47°, 245.53°

Usually, the best way of solving these is to think of it as a quadratic. So, let y = cos(x)

==> y^2 = 2y + 1
==> y^2 - 2y -1 = 0
==> y = (2 +/- sqrt(4 + 4))/2 = 1 +/- sqrt(2)
==> cos(x) = 1 +/- sqrt(2)
cos(x)'s values range from -1 to 1, so we can ignore 1 + sqrt(2) as a solution
So, cos(x) = 1 - sqrt(2)
0r cos(x) = -0.4142
So, x = (approx) 114 degrees or 246 degrees.