First, we find the derivative of the function $f(x) = \dfrac{4 \sin x - 2x - x \cos x}{2 + \cos x}$.

Using the quotient rule,

$f'(x) = \frac{\cos x \, (4 - \cos x)}{(2 + \cos x)^2}.$

Since $(4 - \cos x) > 0$ and $(2 + \cos x)^2 > 0$ for all $x$, the sign of $f'(x)$ depends entirely on $\cos x$.

Therefore, $f'(x) > 0$ when $\cos x > 0$, which is in $(0, \pi/2) \cup (3\pi/2, 2\pi)$, making $f$ increasing there.

Conversely, $f'(x) < 0$ when $\cos x < 0$, which is in $(\pi/2, 3\pi/2)$, making $f$ decreasing there.