To solve this trigonometric equation, we can use the double angle formula for cosine which states that cos$2x$ = 2cos^2$x$ - 1.

Substitute this formula into the equation:

4$2co{s}^2(x)-1$ - 8cos$x$ - 1 = 0
8cos^2$x$ - 4 - 8cos$x$ - 1 = 0
8cos^2$x$ - 8cos$x$ - 5 = 0

Now, let y = cos$x$. The equation becomes:

8y^2 - 8y - 5 = 0

Now we have a quadratic equation that we can solve using the quadratic formula:

y = $$8\pm \surd (8^2-4<em>8</em>(-5))$$ / 2*8
y = $$8\pm \surd (64+160)$$ / 16
y = $$8\pm \surd 224$$ / 16

y = $$8\pm 14.97$$ / 16

Now we solve for y:

y1 = $8+14.97$ / 16 ≈ 1.187
y2 = $8-14.97$ / 16 ≈ -0.874

Since y = cos$x$, we can find the corresponding values of x:

x1 = cos^$-1$$1.187$ is not valid
x2 = cos^$-1$$-0.874$ ≈ 2.610

Therefore, the solutions to the equation 4cos$2x$ - 8cos$x$ - 1 = 0 are x ≈ 2.610 + 2πn, n ∈ ℤ.