If `sqrt 3 sin theta - cos theta = sqrt2`, then `theta`=

Options

A. `2n pi - pi/3 pm (3pi)/4`
B. `n pi + (-1)^n pi/4 + pi/6`
C. `2n pi + pi/3 pm (3pi)/4`
D. `n pi + (-1)^n pi/4 - pi/6`

Answer

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

### Step 1: Rearranging the Equation
We start with the equation:
\[ \sqrt{3} \sin \theta - \cos \theta = \sqrt{2} \]
We can rearrange this equation to isolate the trigonometric terms:
\[ \sqrt{3} \sin \theta = \cos \theta + \sqrt{2} \]

### Step 2: Dividing by 2
Next, we divide both sides of the equation by 2:
\[ \frac{\sqrt{3}}{2} \sin \theta - \frac{1}{2} \cos \theta = \frac{\sqrt{2}}{2} \]

### Step 3: Recognizing Trigonometric Values
We know 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 equation as:
\[ \sin \frac{\pi}{3} \sin \theta - \cos \frac{\pi}{3} \cos \theta = \frac{\sqrt{2}}{2} \]

### Step 4: Using the Sine Difference Formula
Using the sine difference formula:
\[ \sin A \sin B - \cos A \cos B = -\cos(A + B) \]
we can rewrite the left-hand side:
\[ -\cos\left(\theta + \frac{\pi}{3}\right) = \frac{\sqrt{2}}{2} \]

### Step 5: Solving for Cosine
This implies:
\[ \cos\left(\theta + \frac{\pi}{3}\right) = -\frac{\sqrt{2}}{2} \]

### Step 6: Finding Angles
The cosine function equals \(-\frac{\sqrt{2}}{2}\) at angles:
\[ \theta + \frac{\pi}{3} = \frac{3\pi}{4} + 2n\pi \quad \text{or} \quad \theta + \frac{\pi}{3} = \frac{5\pi}{4} + 2n\pi \]
where \(n\) is any integer.

### Step 7: Isolating \(\theta\)
Now, we solve for \(\theta\):
1. From \( \theta + \frac{\pi}{3} = \frac{3\pi}{4} + 2n\pi \):
\[ \theta = \frac{3\pi}{4} - \frac{\pi}{3} + 2n\pi \]
Finding a common denominator (12):
\[ \theta = \frac{9\pi}{12} - \frac{4\pi}{12} + 2n\pi = \frac{5\pi}{12} + 2n\pi \]

2. From \( \theta + \frac{\pi}{3} = \frac{5\pi}{4} + 2n\pi \):
\[ \theta = \frac{5\pi}{4} - \frac{\pi}{3} + 2n\pi \]
Again, finding a common denominator (12):
\[ \theta = \frac{15\pi}{12} - \frac{4\pi}{12} + 2n\pi = \frac{11\pi}{12} + 2n\pi \]

### Final Solution
Thus, the general solutions for \(\theta\) are:
\[ \theta = \frac{5\pi}{12} + 2n\pi \quad \text{and} \quad \theta = \frac{11\pi}{12} + 2n\pi \]