Find the principal and general solution of `tanx=sqrt(3)`

Explanation

To solve the equation \( \tan x = \sqrt{3} \), we will find both the principal and general solutions step by step.

### Step 1: Identify the angle corresponding to \( \tan x = \sqrt{3} \)

The value \( \sqrt{3} \) corresponds to a specific angle in the unit circle. We know that:

\[ \tan \frac{\pi}{3} = \sqrt{3} \]

Thus, we can say:

\[ x = \frac{\pi}{3} + n\pi \]

where \( n \) is any integer, because the tangent function has a period of \( \pi \).

### Step 2: Write the general solution

From the above, we can write the general solution as:

\[ x = \frac{\pi}{3} + n\pi \quad (n \in \mathbb{Z}) \]

### Step 3: Find the principal solutions

The principal solutions are the specific solutions that lie within the interval \( [0, 2\pi) \). We will evaluate the general solution for different integer values of \( n \).

1. **For \( n = 0 \)**:
\[ x = \frac{\pi}{3} + 0 \cdot \pi = \frac{\pi}{3} \]
This is within the interval \( [0, 2\pi) \).

2. **For \( n = 1 \)**:
\[ x = \frac{\pi}{3} + 1 \cdot \pi = \frac{\pi}{3} + \pi = \frac{4\pi}{3} \]
This is also within the interval \( [0, 2\pi) \).

3. **For \( n = 2 \)**:
\[ x = \frac{\pi}{3} + 2 \cdot \pi = \frac{\pi}{3} + 2\pi = \frac{7\pi}{3} \]
This value exceeds \( 2\pi \), so we do not consider it.

Thus, the principal solutions are:

\[ x = \frac{\pi}{3}, \quad x = \frac{4\pi}{3} \]

### Summary of Solutions

- **General Solution**:
\[ x = \frac{\pi}{3} + n\pi \quad (n \in \mathbb{Z}) \]
- **Principal Solutions**:
\[ x = \frac{\pi}{3}, \quad x = \frac{4\pi}{3} \]