Rewrite the equation using a single trigonometric function.

The given equation is $2 \cos^2 x - \sin x = 1$.
Using the identity $\cos^2 x = 1 - \sin^2 x$, substitute it into the equation:

Let $y = \sin x$. The equation becomes $2y^2 + y - 1 = 0$.
Factor the quadratic equation:

This gives two possible values for $y$:

So, $\sin x = \frac{1}{2}$ or $\sin x = -1$.

The principal value for $\sin x = \frac{1}{2}$ is $x = \frac{\pi}{6}$.
Since $\sin x$ is positive, $x$ can be in the first or second quadrant.
In the first quadrant: $x_1 = \frac{\pi}{6}$.
In the second quadrant: $x_2 = \pi - \frac{\pi}{6} = \frac{6\pi - \pi}{6} = \frac{5\pi}{6}$.

The value for which $\sin x = -1$ is $x_3 = \frac{3\pi}{2}$.

The values of $x$ in the interval $0 \le x \le 2\pi$ are $\frac{\pi}{6}$, $\frac{5\pi}{6}$, and $\frac{3\pi}{2}$.

The values of $x$ are $\boxed{\frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}}$.