To solve the equation 2cos^3x + 5cos^2x + 2cosx = 0, we can factor out a cosx to simplify the equation:

cosx(2cos^2x + 5cosx + 2) = 0

Now, the equation reduces to finding the solutions for:

2cos^2x + 5cosx + 2 = 0

This is a quadratic equation in terms of cosx. We can solve this by factoring or using the quadratic formula. Let's use the quadratic formula:

cosx = [-b ± √(b^2 - 4ac)] / 2a

In this case, a = 2, b = 5, and c = 2. Plug these values into the formula:

cosx = [-5 ± √(5^2 - 422)] / 2*2
cosx = [-5 ± √(25 - 16)] / 4
cosx = [-5 ± √9] / 4
cosx = (-5 ± 3) / 4

This gives us two possible solutions for cosx:

cosx = (-5 + 3) / 4 = -2 / 4 = -0.5
cosx = (-5 - 3) / 4 = -8 / 4 = -2

Now, we need to find the corresponding values of x by taking the inverse cosine of these solutions:

x = arccos(-0.5) ≈ 2.0944 or x ≈ 4.1888
x = arccos(-2) is not a real solution (cosine function range is -1 to 1)

Therefore, the solutions to the equation 2cos^3x + 5cos^2x + 2cosx = 0 are x ≈ 2.0944 and x ≈ 4.1888.