If `sqrt(3) cdot cos theta + sin theta = sqrt(2)`, then the general value of `theta` is

Options

A. `npi + (-1)^(n) cdot (pi//4)`
B. `(-1)^(n) cdot (pi//4) - (pi//3)`
C. `npi + (pi//4) - (pi//3)`
D. `npi + (-1)^(n) cdot (pi//4) - (pi//3)`

Answer

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

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

### Step 2: Divide the entire equation by 2
To simplify the equation, we divide all terms by 2:
\[ \frac{\sqrt{3}}{2} \cdot \cos \theta + \frac{1}{2} \cdot \sin \theta = \frac{\sqrt{2}}{2} \]

### Step 3: Identify trigonometric values
We can recognize that:
\[ \frac{\sqrt{3}}{2} = \sin \frac{\pi}{3} \quad \text{and} \quad \frac{1}{2} = \cos \frac{\pi}{3} \]
Thus, we can rewrite the left side as:
\[ \sin \frac{\pi}{3} \cdot \cos \theta + \cos \frac{\pi}{3} \cdot \sin \theta = \frac{\sqrt{2}}{2} \]
And we know that \( \frac{\sqrt{2}}{2} = \sin \frac{\pi}{4} \).

### Step 4: Use the sine addition formula
Using the sine addition formula:
\[ \sin(A + B) = \sin A \cdot \cos B + \cos A \cdot \sin B \]
We can rewrite the equation as:
\[ \sin \left( \theta + \frac{\pi}{3} \right) = \sin \frac{\pi}{4} \]

### Step 5: Solve for \( \theta \)
From the equation \( \sin x = \sin y \), we know:
\[ x = n\pi + (-1)^n y \]
Applying this, we have:
\[ \theta + \frac{\pi}{3} = n\pi + (-1)^n \frac{\pi}{4} \]

### Step 6: Isolate \( \theta \)
Now, we isolate \( \theta \):
\[ \theta = n\pi + (-1)^n \frac{\pi}{4} - \frac{\pi}{3} \]

### Step 7: Simplify the expression
To simplify, we need a common denominator:
\[ \theta = n\pi + (-1)^n \frac{\pi}{4} - \frac{4\pi}{12} = n\pi + (-1)^n \frac{3\pi}{12} - \frac{4\pi}{12} \]
\[ \theta = n\pi + \frac{(-1)^n 3\pi - 4\pi}{12} \]
\[ \theta = n\pi + \frac{(-1)^n 3\pi - 4\pi}{12} \]

### Final Result
Thus, the general value of \( \theta \) is:
\[ \theta = n\pi + \frac{(-1)^n \pi}{12} - \frac{\pi}{3} \]