To solve the equation 2cos^(4)(4x) - cos(4x) - 3 = 0, let's make a substitution to simplify the equation.

Let y = cos(4x).

Now, the equation becomes 2y^4 - y - 3 = 0.

To solve this quadratic equation, we can factor it or use the quadratic formula. Let's use the factorization method:

2y^4 - y - 3 = 0
(2y^2 + 3)(y^2 - 1) = 0
(2y^2 + 3)(y + 1)(y - 1) = 0

Now we have three possibilities for y:

1) 2y^2 + 3 = 0
2y^2 = -3
y^2 = -3/2 (which has no real solutions)

2) y + 1 = 0
y = -1

3) y - 1 = 0
y = 1

Now, substitute y back in terms of cos(4x):

For y = -1:
cos(4x) = -1
4x = π + 2πn, where n is an integer
x = π/4 + π/2n

For y = 1:
cos(4x) = 1
4x = 2πn, where n is an integer
x = πn/2

Therefore, the solutions to the equation 2cos^(4)(4x) - cos(4x) - 3 = 0 are
x = π/4 + π/2n or x = πn/2, where n is an integer.