To solve the equation \(4\cos^2 x - 4\sin x - 1 = 0\), let's start by using the Pythagorean identity, \(\cos^2 x = 1 - \sin^2 x\). Substituting this into the equation gives:

\(4(1 - \sin^2 x) - 4\sin x - 1 = 0\).

Simplifying this, we have:

\(4 - 4\sin^2 x - 4\sin x - 1 = 0\).

Combine like terms:

\(-4\sin^2 x - 4\sin x + 3 = 0\).

Multiply the entire equation by \(-1\) to simplify:

\(4\sin^2 x + 4\sin x - 3 = 0\).

Let \(u = \sin x\). Substituting, we have a quadratic equation:

\(4u^2 + 4u - 3 = 0\).

We can solve this quadratic equation using the quadratic formula, \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = 4\), and \(c = -3\):

\[ u = \frac{-4 \pm \sqrt{4^2 - 4 \times 4 \times (-3)}}{2 \times 4}. \]

Simplifying further:

\[ u = \frac{-4 \pm \sqrt{16 + 48}}{8}, \]
\[ u = \frac{-4 \pm \sqrt{64}}{8}, \]
\[ u = \frac{-4 \pm 8}{8}. \]

This gives us two solutions for \(u\):

1. \(u = \frac{-4 + 8}{8} = \frac{4}{8} = \frac{1}{2},\)
2. \(u = \frac{-4 - 8}{8} = \frac{-12}{8} = -\frac{3}{2}.\)

Since \(\sin x\) must be in the range of \([-1, 1]\), only \(u = \frac{1}{2}\) is valid.

Therefore, \(\sin x = \frac{1}{2}\). The general solutions for \(\sin x = \frac{1}{2}\) are:

\[ x = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad x = \frac{5\pi}{6} + 2k\pi, \]

where \(k\) is any integer.