If `sqrt3cos theta + sin theta = sqrt2` then general value of `theta ` is

Options

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

Answer

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

### Step 1: Rewrite the equation
We start with the equation:
\[ \sqrt{3} \cos \theta + \sin \theta = \sqrt{2} \]
To simplify, we can divide the entire equation by 2:
\[ \frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta = \frac{\sqrt{2}}{2} \]

### Step 2: Identify coefficients
Now we can identify the coefficients:
- \( A = \frac{\sqrt{3}}{2} \)
- \( B = \frac{1}{2} \)
- \( R = \sqrt{A^2 + B^2} = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{1} = 1 \)

### Step 3: Find the angle
Next, we find the angle \( \alpha \) such that:
\[ \cos \alpha = \frac{\sqrt{3}}{2}, \quad \sin \alpha = \frac{1}{2} \]
This corresponds to \( \alpha = \frac{\pi}{6} \).

### Step 4: Rewrite using sine addition formula
Using the sine addition formula, we can rewrite the left-hand side:
\[ R \sin(\theta + \alpha) = \frac{\sqrt{2}}{2} \]
Thus, we have:
\[ \sin\left(\theta + \frac{\pi}{6}\right) = \frac{\sqrt{2}}{2} \]

### Step 5: Solve for \( \theta \)
The general solution for \( \sin x = \frac{\sqrt{2}}{2} \) is:
\[ x = n\pi + (-1)^n \frac{\pi}{4} \]
Substituting \( x = \theta + \frac{\pi}{6} \):
\[ \theta + \frac{\pi}{6} = n\pi + (-1)^n \frac{\pi}{4} \]

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

### Step 7: Simplify the expression
To simplify further, we need a common denominator:
\[ \theta = n\pi + (-1)^n \frac{3\pi}{12} - \frac{2\pi}{12} = n\pi + (-1)^n \frac{3\pi}{12} - \frac{2\pi}{12} \]
This gives us:
\[ \theta = n\pi + \frac{(-1)^n \pi}{4} - \frac{\pi}{6} \]

### Step 8: Final general solution
Thus, the general solution for \( \theta \) can be expressed as:
\[ \theta = n\pi + \frac{(-1)^n \pi}{4} - \frac{\pi}{6} \]