The function f defined by `f(x)=(4sinx-2x-xcosx)/(2+cosx),0lexle2pi` is :

Options

A. increasing in `(0,(pi)/(2))` and decreasing in `((3pi)/(2),2pi)`.
B. increasing in `(-(pi)/(2),0)` and decreasing in `((pi)/(2),(3pi)/(2))uu((3pi)/(2),2pi)`
C. increasing in `((3pi)/(2),2pi)` and decreasing in `(0,(pi)/(2))cap((pi)/(2),(3pi)/(2))`.
D. increasing in `(0,(pi)/(2))uu((3pi)/(2),2pi)and` decreasing in `((pi)/(2),(3pi)/(2))`.

Answer

To solve the problem, we need to analyze the function \( f(x) = \frac{4 \sin x - 2x - x \cos x}{2 + \cos x} \) for \( 0 \leq x \leq 2\pi \) to determine where it is increasing and decreasing.

### Step 1: Simplify the function
We start with the function:
\[ f(x) = \frac{4 \sin x - 2x - x \cos x}{2 + \cos x} \]
We can rewrite this as:
\[ f(x) = \frac{4 \sin x}{2 + \cos x} - \frac{2x + x \cos x}{2 + \cos x} \]
This can be simplified to:
\[ f(x) = \frac{4 \sin x}{2 + \cos x} - x \]

### Step 2: Find the derivative \( f'(x) \)
To determine where the function is increasing or decreasing, we need to find the derivative \( f'(x) \). We will use the quotient rule for differentiation:
\[ f'(x) = \frac{(2 + \cos x)(4 \cos x) - (4 \sin x)(-\sin x)}{(2 + \cos x)^2} - 1 \]
This simplifies to:
\[ f'(x) = \frac{(4 \cos x (2 + \cos x) + 4 \sin^2 x)}{(2 + \cos x)^2} - 1 \]

### Step 3: Set the derivative to zero
To find critical points, we set \( f'(x) = 0 \):
\[ 4 \cos x (2 + \cos x) + 4 \sin^2 x = (2 + \cos x)^2 \]
This equation will help us find where the function changes from increasing to decreasing or vice versa.

### Step 4: Analyze the sign of \( f'(x) \)
We need to analyze the sign of \( f'(x) \) in the intervals determined by the critical points. The function \( f'(x) \) will be positive where \( f(x) \) is increasing and negative where \( f(x) \) is decreasing.

### Step 5: Determine intervals of increase and decrease
1. **Increasing**: When \( f'(x) > 0 \)
- This occurs when \( \cos x \) is positive, which is in the intervals \( [0, \frac{\pi}{2}] \) and \( [\frac{3\pi}{2}, 2\pi] \).

2. **Decreasing**: When \( f'(x) < 0 \)
- This occurs when \( \cos x \) is negative, which is in the intervals \( [\frac{\pi}{2}, \frac{3\pi}{2}] \).

### Conclusion
From our analysis, we find:
- \( f(x) \) is increasing in the intervals \( [0, \frac{\pi}{2}] \) and \( [\frac{3\pi}{2}, 2\pi] \).
- \( f(x) \) is decreasing in the interval \( [\frac{\pi}{2}, \frac{3\pi}{2}] \).

Thus, the correct option is:
**Option 4**: Increasing in \( [0, \frac{\pi}{2}] \) and \( [\frac{3\pi}{2}, 2\pi] \), and decreasing in \( [\frac{\pi}{2}, \frac{3\pi}{2}] \).