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The problem is asking to find the values of \( \theta \) for which \( \tan(\theta) = -\frac{1}{\sqrt{3}} \) within the interval \( 0^\circ \leq \theta \leq 360^\circ \).

### Step-by-Step Solution:

1. **Identify the reference angle:**
   We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \). Therefore, the reference angle for \( \tan(\theta) = -\frac{1}{\sqrt{3}} \) is \( 30^\circ \).

2. **Determine the quadrants where \( \tan(\theta) \) is negative:**
   The tangent function is negative in the second and fourth quadrants.

3. **Find the angles:**
   - In the second quadrant: \( \theta = 180^\circ - 30^\circ = 150^\circ \)
   - In the fourth quadrant: \( \theta = 360^\circ - 30^\circ = 330^\circ \)

### Final Answer:
The values of \( \theta \) for which \( \tan(\theta) = -\frac{1}{\sqrt{3}} \) are:
\[ \theta = 150^\circ, 330^\circ \]

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Would you like further clarification on this solution or any additional details? 

Here are five related questions you might find useful:

1. What is the general solution for \( \tan(\theta) = -\frac{1}{\sqrt{3}} \)?
2. How do the signs of trigonometric functions change across different quadrants?
3. What is the relationship between the tangent function and the unit circle?
4. How can we express angles in radians for the same problem?
5. How do we solve trigonometric equations that involve functions other than tangent?

**Tip:** Always check the quadrant when solving trigonometric equations to determine the correct sign of the function!

Find the values of θ for which tan(θ) = -1/√3, 0° ≤ θ ≤ 360°.

Learn how to solve for θ where tan(θ) = -1/√3 within the interval 0° ≤ θ ≤ 360°. Explore key trigonometric concepts like reference angles and quadrant analysis, with detailed step-by-step explanations.

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