To solve the equation 2cos^2(2x) - 7cos(2x) + 5 = 0, we can let y = cos(2x) and rewrite the equation as 2y^2 - 7y + 5 = 0.

Now, we can factor the quadratic equation as follows:

(2y - 5)(y - 1) = 0

Setting each factor to zero gives us:

2y - 5 = 0 or y - 1 = 0

Solving for y in each case:

Case 1: 2y - 5 = 0
2y = 5
y = 5/2

Case 2: y - 1 = 0
y = 1

Now we substitute back y = cos(2x) in terms of x:

cos(2x) = 5/2 or cos(2x) = 1

Since the cosine function has a range of [-1, 1], the equation cos(2x) = 5/2 has no real solutions.

For cos(2x) = 1, we know that the cosine function equals 1 when the angle is 0 degrees or 360 degrees. Therefore, we have:

2x = 0 + 360n, where n is an integer

Dividing by 2 on both sides gives us:

x = 0 + 180n, where n is an integer

So, the solutions to the equation 2cos^2(2x) - 7cos(2x) + 5 = 0 are x = 0 degrees + 180n degrees, where n is an integer.