To solve the equation \( \sec x = -3 \) in the interval \( \pi \leq x \leq 2\pi \), we can proceed as follows:

1. **Rewrite in terms of cosine**:
   Since \( \sec x = \frac{1}{\cos x} \), the equation becomes:
   \[
   \frac{1}{\cos x} = -3
   \]
   which simplifies to:
   \[
   \cos x = -\frac{1}{3}
   \]

2. **Find the reference angle**:
   We calculate the reference angle by finding the arccosine of \( \frac{1}{3} \):
   \[
   \theta = \cos^{-1} \left( \frac{1}{3} \right) \approx 1.231
   \]

3. **Determine angles in the specified interval**:
   Since \( \cos x \) is negative and we are looking in the interval \( \pi \leq x \leq 2\pi \), \( x \) must be in either the third or fourth quadrants.

   - In the third quadrant: \( x = \pi + \theta \)
   - In the fourth quadrant: \( x = 2\pi - \theta \)

4. **Calculate specific solutions**:
   Substitute \( \theta \approx 1.231 \):
   - Third quadrant solution:
     \[
     x = \pi + 1.231 \approx 4.373
     \]
   - Fourth quadrant solution:
     \[
     x = 2\pi - 1.231 \approx 5.052
     \]

5. **Final answer**:
   The solutions to \( \sec x = -3 \) in the interval \( \pi \leq x \leq 2\pi \) are:
   \[
   x \approx 4.373 \quad \text{and} \quad x \approx 5.052
   \]

Would you like further details or have any questions?

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### Follow-up Questions
1. How do we determine the quadrant in which an angle lies for different trigonometric functions?
2. Why do we restrict the range of arccosine to only yield certain values?
3. Can the secant function have any other negative values within this interval?
4. What are the steps for solving similar equations involving cosecant or cotangent?
5. How would we graph \( \sec x = -3 \) over \( \pi \leq x \leq 2\pi \) to visualize the solutions?

**Tip:** Always check the interval of the trigonometric equation to ensure you find all solutions within the specified range.

Explore how to solve the equation sec x = -3 in the interval [π, 2π]. This guide provides a step-by-step solution, referencing the cosine function and the properties of trigonometric functions. Ideal for high school students in Grades 10-12.

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