Question

2cos²x - 3cosx + 1 = 0
Let cosx = Q:
2Q² - 3Q + 1 = 0
(2Q - 1)(Q-1) = 0
Q = 1/2, 1
So cosx = 1/2 or cosx = 1
x = π/3, 5π/3, 0 (these are the solutions in the interval [0, 2π).
If the instructions called for all possible solutions, you would say x = π/3 + 2πk, 5π/3 + 2πk , 0 + 2πk, where k is any integer.