If the sum of all the solutions of the equation ` 3 cot^(2) theta + 10 cot theta + 3 = 0 ` in ` [ 0, 2 pi ] ` is ` k pi` where ` k in I ` , then find the value of k .

Answer

To solve the equation \( 3 \cot^2 \theta + 10 \cot \theta + 3 = 0 \) and find the sum of all solutions in the interval \( [0, 2\pi] \), we will follow these steps:

### Step 1: Rewrite the equation
The given equation is:
\[ 3 \cot^2 \theta + 10 \cot \theta + 3 = 0 \]

### Step 2: Factor the quadratic equation
We can factor this quadratic equation. We look for two numbers that multiply to \( 3 \times 3 = 9 \) and add to \( 10 \). The numbers \( 9 \) and \( 1 \) work:
\[ 3 \cot^2 \theta + 9 \cot \theta + \cot \theta + 3 = 0 \]
Grouping the terms:
\[ (3 \cot^2 \theta + 9 \cot \theta) + (\cot \theta + 3) = 0 \]
Factoring by grouping:
\[ 3 \cot \theta (\cot \theta + 3) + 1 (\cot \theta + 3) = 0 \]
This gives us:
\[ (3 \cot \theta + 1)(\cot \theta + 3) = 0 \]

### Step 3: Solve for cotangent
Setting each factor to zero:
1. \( 3 \cot \theta + 1 = 0 \) leads to:
\[ \cot \theta = -\frac{1}{3} \]
2. \( \cot \theta + 3 = 0 \) leads to:
\[ \cot \theta = -3 \]

### Step 4: Find angles for each solution
**For \( \cot \theta = -\frac{1}{3} \)**:
- In the second quadrant:
\[ \theta = \pi - \cot^{-1}\left(\frac{1}{3}\right) \]
- In the fourth quadrant:
\[ \theta = 2\pi - \cot^{-1}\left(\frac{1}{3}\right) \]

**For \( \cot \theta = -3 \)**:
- In the second quadrant:
\[ \theta = \pi - \cot^{-1}(3) \]
- In the fourth quadrant:
\[ \theta = 2\pi - \cot^{-1}(3) \]

### Step 5: Sum all solutions
Now we need to sum all solutions:
\[ \theta_1 = \pi - \cot^{-1}\left(\frac{1}{3}\right) \]
\[ \theta_2 = 2\pi - \cot^{-1}\left(\frac{1}{3}\right) \]
\[ \theta_3 = \pi - \cot^{-1}(3) \]
\[ \theta_4 = 2\pi - \cot^{-1}(3) \]

Adding these together:
\[ \text{Sum} = \left(\pi - \cot^{-1}\left(\frac{1}{3}\right)\right) + \left(2\pi - \cot^{-1}\left(\frac{1}{3}\right)\right) + \left(\pi - \cot^{-1}(3)\right) + \left(2\pi - \cot^{-1}(3)\right) \]
\[ = 6\pi - 2\cot^{-1}\left(\frac{1}{3}\right) - 2\cot^{-1}(3) \]

Using the identity \( \cot^{-1}(x) + \tan^{-1}(x) = \frac{\pi}{2} \):
\[ \cot^{-1}(3) = \tan^{-1}\left(\frac{1}{3}\right) \]
Thus,
\[ \cot^{-1}(3) + \cot^{-1}\left(\frac{1}{3}\right) = \frac{\pi}{2} \]
So,
\[ 2\cot^{-1}\left(\frac{1}{3}\right) + 2\cot^{-1}(3) = 2 \cdot \frac{\pi}{2} = \pi \]

Therefore, the sum simplifies to:
\[ 6\pi - \pi = 5\pi \]

### Step 6: Find \( k \)
Since the sum of all solutions is \( 5\pi \) and given that this can be expressed as \( k\pi \), we have:
\[ k = 5 \]

### Final Answer
The value of \( k \) is:
\[ \boxed{5} \]