https://file.aiquickdraw.com/mathbot/file/17264618637843zfua67e.jpg

The problem asks to find two angles between 0 and \( 2\pi \) where \( \tan \theta = -\sqrt{3} \).

### Step-by-step approach:

1. **Standard angle for \( \tan \theta = \sqrt{3} \)**:
   - First, we recognize that \( \tan \theta = \sqrt{3} \) occurs at \( \theta = \frac{\pi}{3} \).

2. **Adjusting for the negative sign**:
   - Since \( \tan \theta = -\sqrt{3} \), we need to find angles where the tangent is negative. Tangent is negative in the second and fourth quadrants.

3. **Finding the solutions**:
   - In the second quadrant, the reference angle is \( \frac{\pi}{3} \), so the solution is \( \theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \).
   - In the fourth quadrant, the reference angle is also \( \frac{\pi}{3} \), so the solution is \( \theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3} \).

### Final Answer:
The two angles where \( \tan \theta = -\sqrt{3} \) are:
\[
\boxed{\frac{2\pi}{3} \text{ and } \frac{5\pi}{3}}
\]

In the image you provided, it seems like the answers written are \( \frac{\pi}{3} \) and \( \frac{4\pi}{3} \), which correspond to where \( \tan \theta = \sqrt{3} \), but for \( -\sqrt{3} \), the correct solutions are \( \frac{2\pi}{3} \) and \( \frac{5\pi}{3} \).

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### Related Questions:
1. Why is tangent positive in the first and third quadrants?
2. How do we calculate reference angles in different quadrants?
3. What is the periodicity of the tangent function?
4. How does one determine where trigonometric functions are positive or negative?
5. What are the basic trigonometric identities related to tangent?

**Tip**: Always check the quadrant when dealing with negative trigonometric values!

Find two angles between 0 and 2π for the given condition. Express your answer in terms of radians. Do not use decimals in your answer. tanθ = -√3

Solve for two angles between 0 and 2π where the tangent of the angle equals -√3. Explore how to use reference angles and quadrants to determine the correct angles, including step-by-step solutions.

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