If `sqrt3cos theta + sin theta = 1 " for " -2pi lt theta lt 2pi`, then `theta =`

Options

A. `-(3pi)/2`
B. `-pi/6`
C. `pi/2`
D. `(11pi)/6`

Answer

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

### Step 1: Rewrite the equation
We start with the equation:
\[ \sqrt{3} \cos \theta + \sin \theta = 1 \]

### Step 2: Divide the equation by 2
To simplify the equation, we can divide everything by 2:
\[ \frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta = \frac{1}{2} \]

### Step 3: Use the cosine addition formula
Recognizing that \( \frac{\sqrt{3}}{2} = \cos \frac{\pi}{6} \) and \( \frac{1}{2} = \sin \frac{\pi}{6} \), we can rewrite the left-hand side using the cosine addition formula:
\[ \cos \left( \theta - \frac{\pi}{6} \right) = \frac{1}{2} \]

### Step 4: Solve for \( \theta - \frac{\pi}{6} \)
Now we need to find the angles where the cosine equals \( \frac{1}{2} \):
\[ \theta - \frac{\pi}{6} = 2n\pi \pm \frac{\pi}{3}, \quad n \in \mathbb{Z} \]

### Step 5: Find the general solutions for \( \theta \)
This gives us two sets of equations:
1. \( \theta - \frac{\pi}{6} = 2n\pi + \frac{\pi}{3} \)
2. \( \theta - \frac{\pi}{6} = 2n\pi - \frac{\pi}{3} \)

From these, we can solve for \( \theta \):
1. \( \theta = 2n\pi + \frac{\pi}{3} + \frac{\pi}{6} = 2n\pi + \frac{\pi}{2} \)
2. \( \theta = 2n\pi - \frac{\pi}{3} + \frac{\pi}{6} = 2n\pi - \frac{\pi}{6} \)

### Step 6: Substitute values of \( n \)
Now we will substitute \( n = -1, 0, 1 \) to find all possible values of \( \theta \) within the interval \( -2\pi < \theta < 2\pi \).

#### For \( n = -1 \):
1. \( \theta = -2\pi + \frac{\pi}{2} = -\frac{3\pi}{2} \)
2. \( \theta = -2\pi - \frac{\pi}{6} = -\frac{13\pi}{6} \) (not valid as it is less than -2π)

#### For \( n = 0 \):
1. \( \theta = 0 + \frac{\pi}{2} = \frac{\pi}{2} \)
2. \( \theta = 0 - \frac{\pi}{6} = -\frac{\pi}{6} \)

#### For \( n = 1 \):
1. \( \theta = 2\pi + \frac{\pi}{2} = \frac{5\pi}{2} \) (not valid as it is greater than 2π)
2. \( \theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6} \)

### Step 7: Collect valid solutions
The valid solutions for \( \theta \) in the interval \( -2\pi < \theta < 2\pi \) are:
1. \( -\frac{3\pi}{2} \)
2. \( \frac{\pi}{2} \)
3. \( -\frac{\pi}{6} \)
4. \( \frac{11\pi}{6} \)

### Final Answer
Thus, the solutions for \( \theta \) are:
\[ \theta = -\frac{3\pi}{2}, \frac{\pi}{2}, -\frac{\pi}{6}, \frac{11\pi}{6} \]