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

Explanation

To find the number of solutions of the equation \( \cot^2(\sin x + 3) = 1 \) in the interval \([0, 3\pi]\), we can follow these steps:

### Step 1: Rewrite the Equation
The equation \( \cot^2(\sin x + 3) = 1 \) can be rewritten as:
\[ \cot(\sin x + 3) = \pm 1 \]
This implies that:
\[ \sin x + 3 = n\pi + \frac{\pi}{4} \quad \text{or} \quad \sin x + 3 = n\pi - \frac{\pi}{4} \]
for some integer \( n \).

### Step 2: Solve for \( \sin x \)
From the above equations, we can express \( \sin x \):
1. \( \sin x = n\pi + \frac{\pi}{4} - 3 \)
2. \( \sin x = n\pi - \frac{\pi}{4} - 3 \)

### Step 3: Determine the Range of \( \sin x \)
The sine function has a range of \([-1, 1]\). Therefore, we need to ensure that:
\[ -1 \leq n\pi + \frac{\pi}{4} - 3 \leq 1 \]
and
\[ -1 \leq n\pi - \frac{\pi}{4} - 3 \leq 1 \]

### Step 4: Analyze Each Case
#### Case 1: \( \sin x = n\pi + \frac{\pi}{4} - 3 \)
1. For \( n = 0 \):
\[ \sin x = 0 + \frac{\pi}{4} - 3 = \frac{\pi}{4} - 3 \approx 0.785 - 3 \approx -2.215 \quad (\text{not valid}) \]
2. For \( n = 1 \):
\[ \sin x = \pi + \frac{\pi}{4} - 3 = 3.14 + 0.785 - 3 \approx 1.925 \quad (\text{not valid}) \]
3. For \( n = 2 \):
\[ \sin x = 2\pi + \frac{\pi}{4} - 3 = 6.28 + 0.785 - 3 \approx 4.065 \quad (\text{not valid}) \]

#### Case 2: \( \sin x = n\pi - \frac{\pi}{4} - 3 \)
1. For \( n = 0 \):
\[ \sin x = 0 - \frac{\pi}{4} - 3 = -\frac{\pi}{4} - 3 \approx -0.785 - 3 \approx -3.785 \quad (\text{not valid}) \]
2. For \( n = 1 \):
\[ \sin x = \pi - \frac{\pi}{4} - 3 = 3.14 - 0.785 - 3 \approx -0.645 \quad (\text{valid}) \]
3. For \( n = 2 \):
\[ \sin x = 2\pi - \frac{\pi}{4} - 3 = 6.28 - 0.785 - 3 \approx 2.495 \quad (\text{not valid}) \]

### Step 5: Find Solutions for Valid Cases
From the valid case \( \sin x = -0.645 \):
1. The solutions for \( \sin x = -0.645 \) in the interval \([0, 3\pi]\) can be found using:
\[ x = \arcsin(-0.645) + 2k\pi \quad \text{and} \quad x = \pi - \arcsin(-0.645) + 2k\pi \]
for integers \( k \).

2. For \( k = 0 \):
- \( x_1 = \arcsin(-0.645) \) (in the fourth quadrant)
- \( x_2 = \pi - \arcsin(-0.645) \) (in the second quadrant)

3. For \( k = 1 \):
- \( x_3 = \arcsin(-0.645) + 2\pi \)
- \( x_4 = \pi - \arcsin(-0.645) + 2\pi \)

### Step 6: Count the Solutions
In the interval \([0, 3\pi]\), we have:
- Two solutions from \( k = 0 \)
- Two solutions from \( k = 1 \)

Thus, the total number of solutions is \( 4 \).

### Final Answer
The number of solutions of the equation \( \cot^2(\sin x + 3) = 1 \) in the interval \([0, 3\pi]\) is **4**.

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