If `sqrt3 sectheta+2=0` , then principal value of `theta` may be

Options

A. `(5pi)/6`
B. `-(5pi)/6`
C. `(7pi)/6`
D. `-(7pi)/6`

Answer

To solve the equation \( \sqrt{3} \sec \theta + 2 = 0 \), we will follow these steps:

### Step 1: Isolate the secant function
Starting from the equation:
\[ \sqrt{3} \sec \theta + 2 = 0 \]
we can isolate \( \sec \theta \):
\[ \sqrt{3} \sec \theta = -2 \]
Dividing both sides by \( \sqrt{3} \):
\[ \sec \theta = -\frac{2}{\sqrt{3}} \]

### Step 2: Convert secant to cosine
Recall that \( \sec \theta = \frac{1}{\cos \theta} \). Therefore, we can rewrite the equation as:
\[ \frac{1}{\cos \theta} = -\frac{2}{\sqrt{3}} \]
Taking the reciprocal gives:
\[ \cos \theta = -\frac{\sqrt{3}}{2} \]

### Step 3: Determine the angles for cosine
The cosine function is negative in the second and third quadrants. The reference angle where \( \cos \theta = \frac{\sqrt{3}}{2} \) is \( \frac{\pi}{6} \). Thus, the angles where \( \cos \theta = -\frac{\sqrt{3}}{2} \) are:
1. In the second quadrant:
\[ \theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \]
2. In the third quadrant:
\[ \theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \]

### Step 4: Principal values
The principal values of \( \theta \) that satisfy \( \cos \theta = -\frac{\sqrt{3}}{2} \) are:
\[ \theta = \frac{5\pi}{6} \quad \text{and} \quad \theta = \frac{7\pi}{6} \]

### Conclusion
Thus, the principal values of \( \theta \) may be \( \frac{5\pi}{6} \) and \( \frac{7\pi}{6} \).