Find all solutions of this equation

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

(cos(x) - 2) * (cos(x) + 1) = 0
cos(x) = -1 , 2

cos(x) cannot equal 2

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

Let use substitution: let a = cos(x)
So cos(^2)x-cosx-2=0 becomes
a^2 - a - 2 = 0
factor
(a - 2)*(a + 1) = 0
a = 2 or -1
since a = cos(x),
cos(x) = 2 or -1.
Since the range of cos(x) is [-1, 1], cos(x) cannot be 2
so cos(x) = -1
thus x = 180 + 360n degrees
or x = π + 2πn radians.
where n is an integer.

cos²x-cosx-2=0
cos²x - cosx + 1/4 = 9/4
(cosx - 1/2)² = 9/4
cosx = 1/2 ± 3/2 = -1, 2
Discard cosx = 2 since |cosx| ≤ 1

cosx = -1
x = arccos(-1) = π ± 2nπ = π(1±2n) : n = any integer.

(cos x-2)(cos x +1)=0
cos x=2 noanswers because -1 cos x=-1
x=180 deg
God bless you.