Solution set of `4 sintheta*costheta-2 costheta-2sqrt3 sin theta+sqrt3 = 0` in the interval `(0, 2 pi)` is

Options

A. `{(3pi)/(4), (7pi)/(4)}`
B. `{(pi)/(3), (5pi)/(3)}`
C. `{(3pi)/(4), pi, (pi)/(3), (5pi)/(3)}`
D. `{(pi)/(6), (5pi)/(6), (11pi)/(6)}`

Answer

To solve the equation \( 4\sin\theta \cos\theta - 2\cos\theta - 2\sqrt{3}\sin\theta - \sqrt{3} = 0 \) in the interval \( (0, 2\pi) \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 4\sin\theta \cos\theta - 2\cos\theta - 2\sqrt{3}\sin\theta - \sqrt{3} = 0 \]

### Step 2: Factor the equation
We can group the terms to factor them. First, we can take \( 2\cos\theta \) common from the first two terms and \( \sqrt{3} \) from the last two terms:
\[ 2\cos\theta(2\sin\theta - 1) - \sqrt{3}(2\sin\theta + 1) = 0 \]

### Step 3: Set the factors to zero
Now we can set each factor to zero:
1. \( 2\sin\theta - 1 = 0 \)
2. \( 2\cos\theta - \sqrt{3} = 0 \)

### Step 4: Solve for \( \sin\theta \)
From the first equation:
\[ 2\sin\theta - 1 = 0 \implies \sin\theta = \frac{1}{2} \]
The solutions for \( \sin\theta = \frac{1}{2} \) in the interval \( (0, 2\pi) \) are:
\[ \theta = \frac{\pi}{6}, \quad \frac{5\pi}{6} \]

### Step 5: Solve for \( \cos\theta \)
From the second equation:
\[ 2\cos\theta - \sqrt{3} = 0 \implies \cos\theta = \frac{\sqrt{3}}{2} \]
The solutions for \( \cos\theta = \frac{\sqrt{3}}{2} \) in the interval \( (0, 2\pi) \) are:
\[ \theta = \frac{\pi}{6}, \quad \frac{11\pi}{6} \]

### Step 6: Combine the solutions
Now we combine all the solutions we found:
- From \( \sin\theta = \frac{1}{2} \): \( \frac{\pi}{6}, \frac{5\pi}{6} \)
- From \( \cos\theta = \frac{\sqrt{3}}{2} \): \( \frac{\pi}{6}, \frac{11\pi}{6} \)

Thus, the complete solution set in the interval \( (0, 2\pi) \) is:
\[ \theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{11\pi}{6} \]

### Final Answer
The solution set of the equation in the interval \( (0, 2\pi) \) is:
\[ \left\{ \frac{\pi}{6}, \frac{5\pi}{6}, \frac{11\pi}{6} \right\} \]