Solving trig equation

need a little help solving this trig equation, not too sure what to do, i lost my notes on trig identities

cos2x + cosx = 0

That should be cos(2x). The parentheses avoid any possible ambiguity

One of the double angle formulas for cosine is cos(2x) = 2cos²(x) – 1

cos(2x) + cos(x) = 2cos²(x) + cos(x) – 1 = (2cos(x) – 1)(cos(x) + 1) = 0

2cos(x) – 1 = 0 ⇒ cos(x) = 1/2
x = π/3 + 2kπ or 5π/3 + 2kπ where k is any integer

cos(x) + 1 = 0 ⇒ cos(x) = –1
x = π + 2kπ

The complete solution set is:
x = π/3 + 2kπ
x = π + 2kπ
x = 5π/3 + 2kπ

cos^2 x + cos x = 0
cos x (cos x + 1) = 0
cos x = 0 or cos x + 1 = 0
x = π/2, 3π/2......cos x = -1
.............................x = π