If `sqrt(3)sintheta-costheta=0`, then one general value of `theta` is:

Options

A. `(npi)/(2)`
B. `(npi)/(4)`
C. `(npi)/(6)`
D. `(npi)/(3)`

Answer

To solve the equation \( \sqrt{3} \sin \theta - \cos \theta = 0 \), we will follow these steps:

### Step 1: Rearrange the equation
We start with the given equation:
\[ \sqrt{3} \sin \theta - \cos \theta = 0 \]
Rearranging this gives:
\[ \sqrt{3} \sin \theta = \cos \theta \]

### Step 2: Divide both sides by \(\cos \theta\)
Assuming \(\cos \theta \neq 0\), we divide both sides by \(\cos \theta\):
\[ \frac{\sqrt{3} \sin \theta}{\cos \theta} = 1 \]
This simplifies to:
\[ \sqrt{3} \tan \theta = 1 \]

### Step 3: Solve for \(\tan \theta\)
Now we can isolate \(\tan \theta\):
\[ \tan \theta = \frac{1}{\sqrt{3}} \]

### Step 4: Find the angle \(\theta\)
We know that \(\tan \theta = \frac{1}{\sqrt{3}}\) corresponds to specific angles. The principal value for \(\theta\) is:
\[ \theta = \frac{\pi}{6} \]

### Step 5: Write the general solution
The general solution for \(\tan \theta = \tan \alpha\) is given by:
\[ \theta = n\pi + \alpha \]
where \(n\) is any integer. Substituting \(\alpha = \frac{\pi}{6}\), we get:
\[ \theta = n\pi + \frac{\pi}{6} \]

### Final Answer
Thus, one general value of \(\theta\) is:
\[ \theta = n\pi + \frac{\pi}{6} \]

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