The sum of the roots of the equation `|sqrt3cos x-sinx|=2" in "[0, 4pi]` is `kpi`, then the value of 6k is

Answer

To solve the equation \( |\sqrt{3} \cos x - \sin x| = 2 \) in the interval \( [0, 4\pi] \), we can follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:
\[ |\sqrt{3} \cos x - \sin x| = 2 \]
This can be split into two cases:
1. \( \sqrt{3} \cos x - \sin x = 2 \)
2. \( \sqrt{3} \cos x - \sin x = -2 \)

### Step 2: Solve the First Case
For the first case:
\[ \sqrt{3} \cos x - \sin x = 2 \]
Rearranging gives:
\[ \sqrt{3} \cos x = \sin x + 2 \]
Dividing both sides by 2:
\[ \frac{\sqrt{3}}{2} \cos x = \frac{1}{2} \sin x + 1 \]
Recognizing that \( \frac{\sqrt{3}}{2} = \cos \frac{\pi}{6} \) and \( \frac{1}{2} = \sin \frac{\pi}{6} \), we can rewrite this as:
\[ \cos \frac{\pi}{6} \cos x - \sin \frac{\pi}{6} \sin x = 1 \]
Using the cosine addition formula:
\[ \cos\left(x + \frac{\pi}{6}\right) = 1 \]
This implies:
\[ x + \frac{\pi}{6} = 2n\pi \quad \text{for } n \in \mathbb{Z} \]
Thus:
\[ x = 2n\pi - \frac{\pi}{6} \]

### Step 3: Find Valid Solutions for \( n \)
We need to find \( n \) such that \( x \) lies in the interval \( [0, 4\pi] \):
- For \( n = 0 \): \( x = -\frac{\pi}{6} \) (not valid)
- For \( n = 1 \): \( x = 2\pi - \frac{\pi}{6} = \frac{12\pi}{6} - \frac{\pi}{6} = \frac{11\pi}{6} \)
- For \( n = 2 \): \( x = 4\pi - \frac{\pi}{6} = \frac{24\pi}{6} - \frac{\pi}{6} = \frac{23\pi}{6} \)

### Step 4: Solve the Second Case
For the second case:
\[ \sqrt{3} \cos x - \sin x = -2 \]
Rearranging gives:
\[ \sqrt{3} \cos x = \sin x - 2 \]
Dividing by 2:
\[ \frac{\sqrt{3}}{2} \cos x = \frac{1}{2} \sin x - 1 \]
This can be rewritten as:
\[ \cos \frac{\pi}{6} \cos x - \sin \frac{\pi}{6} \sin x = -1 \]
Using the cosine addition formula:
\[ \cos\left(x + \frac{\pi}{6}\right) = -1 \]
This implies:
\[ x + \frac{\pi}{6} = (2n + 1)\pi \quad \text{for } n \in \mathbb{Z} \]
Thus:
\[ x = (2n + 1)\pi - \frac{\pi}{6} \]

### Step 5: Find Valid Solutions for \( n \)
We need to find \( n \) such that \( x \) lies in the interval \( [0, 4\pi] \):
- For \( n = 0 \): \( x = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \)
- For \( n = 1 \): \( x = 3\pi - \frac{\pi}{6} = \frac{17\pi}{6} \)
- For \( n = 2 \): \( x = 5\pi - \frac{\pi}{6} = \frac{29\pi}{6} \) (not valid since \( \frac{29\pi}{6} > 4\pi \))

### Step 6: Collect All Valid Solutions
The valid solutions in the interval \( [0, 4\pi] \) are:
1. From the first case: \( \frac{11\pi}{6}, \frac{23\pi}{6} \)
2. From the second case: \( \frac{5\pi}{6}, \frac{17\pi}{6} \)

### Step 7: Calculate the Sum of the Roots
Now, we sum the valid solutions:
\[ \frac{11\pi}{6} + \frac{23\pi}{6} + \frac{5\pi}{6} + \frac{17\pi}{6} = \frac{11 + 23 + 5 + 17}{6} \pi = \frac{56\pi}{6} = \frac{28\pi}{3} \]
Thus, we have:
\[ k\pi = \frac{28\pi}{3} \implies k = \frac{28}{3} \]

### Step 8: Find the Value of \( 6k \)
Now, we calculate:
\[ 6k = 6 \times \frac{28}{3} = 56 \]

### Final Answer
The value of \( 6k \) is:
\[ \boxed{56} \]