To solve this equation, we first need to factorize the quadratic equation (2·cos² x - 3·cos x - 2) = 0.

Let's denote cos x as t.

Then the equation becomes 2t² - 3t - 2 = 0

Now, we need to factorize the quadratic equation:

2t² - 3t - 2 = (2t + 1)(t - 2) = 0

This gives two solutions: t = -1/2 and t = 2.

Since t = cos x, we have two cases to consider:

Case 1: cos x = -1/2
In this case, x = π/3 + 2πn or x = 5π/3 + 2πn, where n is an integer.

Case 2: cos x = 2
This case is not valid because the range of cosine function is [-1, 1].

Therefore, the solutions to the equation are x = π/3 + 2πn or x = 5π/3 + 2πn, where n is an integer.