The general solution of equation `sin^2 theta sec theta + sqrt3 tan theta = 0` is

Options

A. `theta = npi + (-1)^(n + 1) pi/3`
B. `theta = n pi`
C. `theta = n pi + (-1)^(n +1) pi/6`
D. `theta = (n pi)/2`

Answer

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

### Step 1: Rewrite the Equation
We know that:
\[ \sec \theta = \frac{1}{\cos \theta} \quad \text{and} \quad \tan \theta = \frac{\sin \theta}{\cos \theta} \]
Substituting these into the equation gives:
\[ \sin^2 \theta \cdot \frac{1}{\cos \theta} + \sqrt{3} \cdot \frac{\sin \theta}{\cos \theta} = 0 \]
This simplifies to:
\[ \frac{\sin^2 \theta + \sqrt{3} \sin \theta}{\cos \theta} = 0 \]

### Step 2: Set the Numerator to Zero
Since the fraction is equal to zero, we can set the numerator equal to zero:
\[ \sin^2 \theta + \sqrt{3} \sin \theta = 0 \]

### Step 3: Factor the Equation
Factoring out \( \sin \theta \) gives:
\[ \sin \theta (\sin \theta + \sqrt{3}) = 0 \]

### Step 4: Solve Each Factor
Setting each factor to zero gives us two equations:
1. \( \sin \theta = 0 \)
2. \( \sin \theta + \sqrt{3} = 0 \) or \( \sin \theta = -\sqrt{3} \)

### Step 5: Solve \( \sin \theta = 0 \)
The general solution for \( \sin \theta = 0 \) is:
\[ \theta = n\pi \quad (n \in \mathbb{Z}) \]

### Step 6: Solve \( \sin \theta = -\sqrt{3} \)
However, \( \sin \theta \) must lie between -1 and 1. Since \( -\sqrt{3} \) is less than -1, this case does not yield any valid solutions.

### Final Solution
Thus, the only valid solutions come from \( \sin \theta = 0 \):
\[ \theta = n\pi \quad (n \in \mathbb{Z}) \]