`f(x) =4 cos^3 x + 3 sqrt 3 cos^2 x – 1` then, find number of the local maxima in internal `[0, 2 pi]`

Answer

To find the number of local maxima of the function \( f(x) = 4 \cos^3 x + 3 \sqrt{3} \cos^2 x - 1 \) in the interval \( [0, 2\pi] \), we will follow these steps:

### Step 1: Differentiate the Function
We need to find the first derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx}(4 \cos^3 x + 3 \sqrt{3} \cos^2 x - 1) \]
Using the chain rule and product rule, we differentiate each term:
- The derivative of \( 4 \cos^3 x \) is \( 12 \cos^2 x (-\sin x) = -12 \cos^2 x \sin x \).
- The derivative of \( 3 \sqrt{3} \cos^2 x \) is \( 6 \sqrt{3} \cos x (-\sin x) = -6 \sqrt{3} \cos x \sin x \).
- The derivative of the constant \( -1 \) is \( 0 \).

Thus, we have:
\[ f'(x) = -12 \cos^2 x \sin x - 6 \sqrt{3} \cos x \sin x \]
Factoring out \( -6 \sin x \cos x \):
\[ f'(x) = -6 \sin x \cos x (2 \cos x + \sqrt{3}) \]

### Step 2: Set the Derivative to Zero
To find the critical points, we set \( f'(x) = 0 \):
\[ -6 \sin x \cos x (2 \cos x + \sqrt{3}) = 0 \]
This gives us three cases to consider:
1. \( \sin x = 0 \)
2. \( \cos x = 0 \)
3. \( 2 \cos x + \sqrt{3} = 0 \)

### Step 3: Solve Each Case
1. **Case 1: \( \sin x = 0 \)**
- \( x = 0, \pi, 2\pi \)

2. **Case 2: \( \cos x = 0 \)**
- \( x = \frac{\pi}{2}, \frac{3\pi}{2} \)

3. **Case 3: \( 2 \cos x + \sqrt{3} = 0 \)**
- \( \cos x = -\frac{\sqrt{3}}{2} \)
- This occurs at:
- \( x = \frac{5\pi}{6} \) (2nd quadrant)
- \( x = \frac{7\pi}{6} \) (3rd quadrant)

### Step 4: List All Critical Points
The critical points in the interval \( [0, 2\pi] \) are:
- \( x = 0 \)
- \( x = \pi \)
- \( x = 2\pi \)
- \( x = \frac{\pi}{2} \)
- \( x = \frac{3\pi}{2} \)
- \( x = \frac{5\pi}{6} \)
- \( x = \frac{7\pi}{6} \)

### Step 5: Determine Local Maxima
To determine where the local maxima occur, we analyze the sign of \( f'(x) \) around the critical points:

- **Intervals to check**:
- \( (0, \frac{\pi}{2}) \)
- \( (\frac{\pi}{2}, \frac{5\pi}{6}) \)
- \( (\frac{5\pi}{6}, \pi) \)
- \( (\pi, \frac{7\pi}{6}) \)
- \( (\frac{7\pi}{6}, \frac{3\pi}{2}) \)
- \( (\frac{3\pi}{2}, 2\pi) \)

By testing points in these intervals, we find:
- \( f'(x) \) changes from positive to negative at \( x = \frac{5\pi}{6} \) (local maximum).
- \( f'(x) \) changes from positive to negative at \( x = \frac{7\pi}{6} \) (local maximum).

### Conclusion
The local maxima occur at \( x = \frac{5\pi}{6} \) and \( x = \frac{7\pi}{6} \). Therefore, the number of local maxima in the interval \( [0, 2\pi] \) is:
\[ \boxed{2} \]