Solve the equation for ` 0 le x le 2pi.`
`tan theta + sqrt3 =0`

Answer

To solve the equation \( \tan \theta + \sqrt{3} = 0 \) for \( 0 \leq \theta \leq 2\pi \), we can follow these steps:

### Step 1: Rearrange the equation
We start with the equation:
\[ \tan \theta + \sqrt{3} = 0 \]
Rearranging gives us:
\[ \tan \theta = -\sqrt{3} \]

### Step 2: Identify the reference angle
We know that \( \tan \theta = \sqrt{3} \) at the reference angle \( \theta = \frac{\pi}{3} \). However, since we need \( \tan \theta = -\sqrt{3} \), we will look for angles where the tangent function is negative.

### Step 3: Determine the quadrants
The tangent function is negative in the second and fourth quadrants.

### Step 4: Find the angles in the specified quadrants
1. **Second Quadrant**:
The angle can be found using:
\[ \theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]

2. **Fourth Quadrant**:
The angle can be found using:
\[ \theta = 2\pi - \frac{\pi}{3} = 2\pi - \frac{1\pi}{3} = \frac{6\pi}{3} - \frac{1\pi}{3} = \frac{5\pi}{3} \]

### Step 5: Write the final solutions
Thus, the solutions for the equation \( \tan \theta + \sqrt{3} = 0 \) in the interval \( 0 \leq \theta \leq 2\pi \) are:
\[ \theta = \frac{2\pi}{3}, \quad \theta = \frac{5\pi}{3} \]

### Summary of Solutions
The final solutions are:
\[ \theta = \frac{2\pi}{3}, \quad \theta = \frac{5\pi}{3} \]