Solve the following inequality: `arc cot^(2)x-5 arc cotx+6gt0`

Explanation

To solve the inequality \( \text{arc cot}^2 x - 5 \text{arc cot} x + 6 > 0 \), we can follow these steps:

### Step 1: Substitute \( t = \text{arc cot} x \)
We let \( t = \text{arc cot} x \). Then the inequality becomes:
\[ t^2 - 5t + 6 > 0 \]

### Step 2: Factor the quadratic expression
Next, we need to factor the quadratic expression \( t^2 - 5t + 6 \):
\[ t^2 - 5t + 6 = (t - 2)(t - 3) \]
Thus, the inequality can be rewritten as:
\[ (t - 2)(t - 3) > 0 \]

### Step 3: Determine the critical points
The critical points from the factored inequality are \( t = 2 \) and \( t = 3 \). We will analyze the sign of the product \( (t - 2)(t - 3) \) in the intervals determined by these points.

### Step 4: Test intervals
We will test the intervals:
1. \( (-\infty, 2) \)
2. \( (2, 3) \)
3. \( (3, \infty) \)

- **Interval \( (-\infty, 2) \)**: Choose \( t = 0 \)
\[ (0 - 2)(0 - 3) = (−2)(−3) = 6 > 0 \]

- **Interval \( (2, 3) \)**: Choose \( t = 2.5 \)
\[ (2.5 - 2)(2.5 - 3) = (0.5)(−0.5) = −0.25 < 0 \]

- **Interval \( (3, \infty) \)**: Choose \( t = 4 \)
\[ (4 - 2)(4 - 3) = (2)(1) = 2 > 0 \]

### Step 5: Combine the results
From our tests, we find that the inequality \( (t - 2)(t - 3) > 0 \) holds for:
\[ t \in (-\infty, 2) \cup (3, \infty) \]

### Step 6: Convert back to \( x \)
Since \( t = \text{arc cot} x \), we need to find the corresponding values of \( x \):
- For \( t < 2 \): \( \text{arc cot} x < 2 \) implies \( x > \cot(2) \)
- For \( t > 3 \): \( \text{arc cot} x > 3 \) implies \( x < \cot(3) \)

### Step 7: Final solution
Thus, the solution set for \( x \) is:
\[ x \in (-\infty, \cot(3)) \cup (\cot(2), \infty) \]