If `sqrt3sin theta+costhetagt0, then theta` lies in the interval

Options

A. `(-pi//3,pi//2)`
B. `(-pi//6,5pi//6)`
C. `(pi//4,pi//3)`
D. none of these

Answer

To solve the inequality \( \sqrt{3} \sin \theta + \cos \theta > 0 \), we will follow these steps:

### Step 1: Rewrite the inequality
We start with the inequality:
\[ \sqrt{3} \sin \theta + \cos \theta > 0 \]
To make it easier to analyze, we can multiply and divide by 2:
\[ 2 \cdot \frac{\sqrt{3}}{2} \sin \theta + 2 \cdot \frac{1}{2} \cos \theta > 0 \]
This simplifies to:
\[ \sqrt{3} \sin \theta + \cos \theta > 0 \]

### Step 2: Use trigonometric identities
We can express \( \frac{\sqrt{3}}{2} \) and \( \frac{1}{2} \) in terms of sine and cosine:
\[ \frac{\sqrt{3}}{2} = \cos \frac{\pi}{6}, \quad \frac{1}{2} = \sin \frac{\pi}{6} \]
Thus, we can rewrite the inequality as:
\[ 2 \cdot \cos \frac{\pi}{6} \sin \theta + 2 \cdot \sin \frac{\pi}{6} \cos \theta > 0 \]
This can be recognized as:
\[ \sin \left( \frac{\pi}{6} + \theta \right) > 0 \]

### Step 3: Determine when the sine function is positive
The sine function is positive in the intervals:
\[ (0, \pi) + 2k\pi \quad \text{for any integer } k \]
Thus, we have:
\[ \frac{\pi}{6} + \theta > 0 \quad \text{and} \quad \frac{\pi}{6} + \theta < \pi \]

### Step 4: Solve the inequalities
1. From \( \frac{\pi}{6} + \theta > 0 \):
\[ \theta > -\frac{\pi}{6} \]

2. From \( \frac{\pi}{6} + \theta < \pi \):
\[ \theta < \pi - \frac{\pi}{6} = \frac{5\pi}{6} \]

### Step 5: Combine the results
Combining these results, we find:
\[ -\frac{\pi}{6} < \theta < \frac{5\pi}{6} \]

### Final Answer
Thus, the interval in which \( \theta \) lies is:
\[ \left( -\frac{\pi}{6}, \frac{5\pi}{6} \right) \]