Step 1: Locate the Critical Points:
Differentiating $f(x)=\dfrac{4\sin x-2x-x\cos x}{2+\cos x}$ by the quotient rule and simplifying (using $\sin^2x+\cos^2x=1$) gives the compact form below.
\[ f'(x)=\frac{\cos x(4-\cos x)}{(2+\cos x)^2} \]
Critical points occur where $f'(x)=0$. Since $4-\cos x$ is never $0$ (as $\cos x\le1<4$) and the denominator is never $0$, the only critical points come from $\cos x=0$.
Step 2: Break the Domain Into Test Intervals:
On $[0,2\pi]$, $\cos x=0$ at $x=\pi/2$ and $x=3\pi/2$, splitting the domain into three pieces.
\[ (0,\pi/2), \qquad (\pi/2,3\pi/2), \qquad (3\pi/2,2\pi) \]
Step 3: Test a Sample Point in Each Piece:
Substitute one convenient value of x from each piece directly into $f'(x)=\dfrac{\cos x(4-\cos x)}{(2+\cos x)^2}$ and check the sign.
At $x=\pi/4$: $\cos(\pi/4)=\tfrac{\sqrt2}{2}>0$, so $f'(\pi/4)>0$.
At $x=\pi$: $\cos\pi=-1<0$, so $f'(\pi)<0$.
At $x=7\pi/4$: $\cos(7\pi/4)=\tfrac{\sqrt2}{2}>0$, so $f'(7\pi/4)>0$.
Step 4: Read Off Monotonicity:
A positive test value means f is strictly increasing on that piece, and a negative test value means f is strictly decreasing there.
So $f$ increases on $(0,\pi/2)$ and $(3\pi/2,2\pi)$, and decreases on $(\pi/2,3\pi/2)$.
Final Answer:
Sample point testing confirms f increases before and after the middle interval, and decreases in between.
\[ \boxed{\text{Increasing on } \left(0,\tfrac{\pi}{2}\right)\cup\left(\tfrac{3\pi}{2},2\pi\right), \quad \text{Decreasing on } \left(\tfrac{\pi}{2},\tfrac{3\pi}{2}\right)} \]