Solve the general vlaue.
`sin theta + sqrt3 cos theta = sqrt2`

Answer

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

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

### Step 2: Multiply and divide by \(\sqrt{1^2 + (\sqrt{3})^2}\)
We can express the left-hand side in a single trigonometric function. The expression can be rewritten by multiplying and dividing by \(\sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2\):
\[ \frac{1}{2} \sin \theta + \frac{\sqrt{3}}{2} \cos \theta = \frac{\sqrt{2}}{2} \]

### Step 3: Identify the angles
Notice that:
\[ \frac{1}{2} = \sin \frac{\pi}{6} \quad \text{and} \quad \frac{\sqrt{3}}{2} = \cos \frac{\pi}{6} \]
Thus, we can rewrite the left-hand side as:
\[ \sin \frac{\pi}{6} \sin \theta + \cos \frac{\pi}{6} \cos \theta \]

### Step 4: Use the sine addition formula
Using the sine addition formula:
\[ \sin A \cos B + \cos A \sin B = \sin(A + B) \]
we can rewrite the equation as:
\[ \sin\left(\theta + \frac{\pi}{6}\right) = \frac{\sqrt{2}}{2} \]

### Step 5: Solve for the angle
The general solution for \(\sin x = \frac{\sqrt{2}}{2}\) is:
\[ x = n\pi + (-1)^n \frac{\pi}{4}, \quad n \in \mathbb{Z} \]
Thus, we have:
\[ \theta + \frac{\pi}{6} = n\pi + (-1)^n \frac{\pi}{4} \]

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

### Step 7: Simplify the expression
To combine the terms, we need a common denominator (which is 12):
\[ \theta = n\pi + (-1)^n \frac{3\pi}{12} - \frac{2\pi}{12} \]
\[ \theta = n\pi + \frac{(-1)^n 3\pi - 2\pi}{12} \]
\[ \theta = n\pi + \frac{(-1)^n \pi + \pi}{12} \]
This gives us the final solution:
\[ \theta = n\pi + \frac{(-1)^n \pi + \pi}{12} \]

### Final Answer
Thus, the general solution for the equation \( \sin \theta + \sqrt{3} \cos \theta = \sqrt{2} \) is:
\[ \theta = n\pi + \frac{(-1)^n \pi + \pi}{12}, \quad n \in \mathbb{Z} \]