Find the number of solution of the equation `cot^(2) (sin x+3)=1` in `[0, 3pi]`.

Explanation

To solve the equation \( \cot^2(\sin x + 3) = 1 \) in the interval \([0, 3\pi]\), we will follow these steps:

### Step 1: Rewrite the Equation
The equation \( \cot^2(\sin x + 3) = 1 \) can be rewritten using the identity \( \cot^2 \theta = 1 \) which implies \( \theta = n\pi \) or \( \theta = n\pi + \frac{\pi}{2} \) for integers \( n \).

### Step 2: Set Up the General Solutions
From \( \cot^2(\sin x + 3) = 1 \), we have:
1. \( \sin x + 3 = n\pi \)
2. \( \sin x + 3 = n\pi + \frac{\pi}{2} \)

### Step 3: Solve for \( \sin x \)
For the first case:
\[ \sin x = n\pi - 3 \]
For the second case:
\[ \sin x = n\pi + \frac{\pi}{2} - 3 \]

### Step 4: Determine the Range of \( n \)
Since \( \sin x \) must be in the range \([-1, 1]\), we will find the possible values of \( n \) for both equations.

#### Case 1: \( \sin x = n\pi - 3 \)
1. Set the inequality:
\[ -1 \leq n\pi - 3 \leq 1 \]
This gives:
\[ 2 \leq n\pi \leq 4 \]
Dividing by \( \pi \):
\[ \frac{2}{\pi} \leq n \leq \frac{4}{\pi} \]
Since \( \frac{2}{\pi} \approx 0.636 \) and \( \frac{4}{\pi} \approx 1.273 \), the possible integer value for \( n \) is \( n = 1 \).

So, for \( n = 1 \):
\[ \sin x = \pi - 3 \]
Since \( \pi - 3 \approx 0.14 \), which is valid.

#### Case 2: \( \sin x = n\pi + \frac{\pi}{2} - 3 \)
1. Set the inequality:
\[ -1 \leq n\pi + \frac{\pi}{2} - 3 \leq 1 \]
This gives:
\[ 2 \leq n\pi + \frac{\pi}{2} \leq 4 \]
Rearranging gives:
\[ 2 - \frac{\pi}{2} \leq n\pi \leq 4 - \frac{\pi}{2} \]
Dividing by \( \pi \):
\[ \frac{2 - \frac{\pi}{2}}{\pi} \leq n \leq \frac{4 - \frac{\pi}{2}}{\pi} \]
Evaluating gives:
\[ \frac{2}{\pi} - \frac{1}{2} \leq n \leq \frac{4}{\pi} - \frac{1}{2} \]
Since both bounds are negative, there are no valid integer values for \( n \).

### Step 5: Count the Solutions
From Case 1, we found one valid solution:
\[ \sin x = \pi - 3 \]
To find \( x \):
\[ x = \arcsin(\pi - 3) \quad \text{and} \quad x = \pi - \arcsin(\pi - 3) \]
Since \( \sin x \) has a periodic nature, we also consider the periodic solutions in the interval \([0, 3\pi]\).

### Conclusion
Thus, the total number of solutions in the interval \([0, 3\pi]\) is 2 (one for each quadrant where sine is positive).