Solve the equation `sqrt3 cos x + sin x =1 ` for `- 2pi lt x le 2pi.`

Answer

To solve the equation \( \sqrt{3} \cos x + \sin x = 1 \) for \( -2\pi < x \leq 2\pi \), we can follow these steps:

### Step 1: Rewrite the Equation
Start with the given equation:
\[ \sqrt{3} \cos x + \sin x = 1 \]

### Step 2: Divide by 2
To simplify the equation, divide all terms by 2:
\[ \frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x = \frac{1}{2} \]

### Step 3: Identify Trigonometric Values
Recognize that:
\[ \frac{\sqrt{3}}{2} = \sin\left(\frac{\pi}{3}\right) \quad \text{and} \quad \frac{1}{2} = \cos\left(\frac{\pi}{3}\right) \]
Thus, we can rewrite the equation as:
\[ \sin\left(\frac{\pi}{3}\right) \cos x + \cos\left(\frac{\pi}{3}\right) \sin x = \frac{1}{2} \]

### Step 4: Apply the Sine Addition Formula
Using the sine addition formula \( \sin(A + B) = \sin A \cos B + \cos A \sin B \):
\[ \sin\left(x + \frac{\pi}{3}\right) = \frac{1}{2} \]

### Step 5: Solve for \( x + \frac{\pi}{3} \)
The general solution for \( \sin A = \sin B \) is:
\[ A = n\pi + (-1)^n B \]
Here, let \( A = x + \frac{\pi}{3} \) and \( B = \frac{\pi}{6} \):
\[ x + \frac{\pi}{3} = n\pi + (-1)^n \frac{\pi}{6} \]

### Step 6: Isolate \( x \)
Rearranging gives:
\[ x = n\pi + (-1)^n \frac{\pi}{6} - \frac{\pi}{3} \]
This simplifies to:
\[ x = n\pi + (-1)^n \frac{\pi}{6} - \frac{2\pi}{6} \]
\[ x = n\pi + (-1)^n \frac{\pi}{6} - \frac{\pi}{3} \]
\[ x = n\pi + \left((-1)^n - 2\right) \frac{\pi}{6} \]

### Step 7: Find Specific Solutions
Now we need to find specific values of \( n \) such that \( -2\pi < x \leq 2\pi \).

1. **For \( n = 0 \)**:
\[ x = 0 + (-1)^0 \frac{\pi}{6} - \frac{\pi}{3} = \frac{\pi}{6} - \frac{2\pi}{6} = -\frac{\pi}{6} \]

2. **For \( n = 1 \)**:
\[ x = 1\pi + (-1)^1 \frac{\pi}{6} - \frac{\pi}{3} = \pi - \frac{\pi}{6} - \frac{2\pi}{6} = \pi - \frac{3\pi}{6} = \pi - \frac{\pi}{2} = \frac{\pi}{2} \]

3. **For \( n = 2 \)**:
\[ x = 2\pi + (-1)^2 \frac{\pi}{6} - \frac{\pi}{3} = 2\pi + \frac{\pi}{6} - \frac{2\pi}{6} = 2\pi - \frac{\pi}{6} = \frac{12\pi}{6} - \frac{\pi}{6} = \frac{11\pi}{6} \]

4. **For \( n = -1 \)**:
\[ x = -\pi + (-1)^{-1} \frac{\pi}{6} - \frac{\pi}{3} = -\pi - \frac{\pi}{6} - \frac{2\pi}{6} = -\pi - \frac{3\pi}{6} = -\pi - \frac{\pi}{2} = -\frac{3\pi}{2} \]

5. **For \( n = -2 \)**:
\[ x = -2\pi + (-1)^{-2} \frac{\pi}{6} - \frac{\pi}{3} = -2\pi + \frac{\pi}{6} - \frac{2\pi}{6} = -2\pi - \frac{\pi}{3} = -\frac{12\pi}{6} - \frac{2\pi}{6} = -\frac{14\pi}{6} \quad \text{(not in range)} \]

### Step 8: Compile Solutions
The valid solutions in the interval \( -2\pi < x \leq 2\pi \) are:
\[ x = -\frac{\pi}{6}, \quad x = \frac{\pi}{2}, \quad x = \frac{11\pi}{6}, \quad x = -\frac{3\pi}{2} \]

### Final Answer
The solutions are:
\[ x = -\frac{3\pi}{2}, -\frac{\pi}{6}, \frac{\pi}{2}, \frac{11\pi}{6} \]