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

Answer

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

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

### Step 2: Identify the Coefficients
We can express \( \sin \theta + \cos \theta \) in a different form. We know that:
\[ \sin \theta + \cos \theta = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta \right) \]
Here, the coefficients of \( \sin \theta \) and \( \cos \theta \) are both \( \frac{1}{\sqrt{2}} \).

### Step 3: Use the Sine Addition Formula
Using the sine addition formula, we can rewrite the expression as:
\[ \sin \theta + \cos \theta = \sqrt{2} \sin\left(\theta + \frac{\pi}{4}\right) \]
Thus, we have:
\[ \sqrt{2} \sin\left(\theta + \frac{\pi}{4}\right) = \sqrt{2} \]

### Step 4: Simplify the Equation
Dividing both sides by \( \sqrt{2} \):
\[ \sin\left(\theta + \frac{\pi}{4}\right) = 1 \]

### Step 5: Solve for \( \theta \)
The general solution for \( \sin x = 1 \) is:
\[ x = \frac{\pi}{2} + 2n\pi \quad \text{for } n \in \mathbb{Z} \]
Substituting back for \( x \):
\[ \theta + \frac{\pi}{4} = \frac{\pi}{2} + 2n\pi \]

### Step 6: Isolate \( \theta \)
Now, we isolate \( \theta \):
\[ \theta = \frac{\pi}{2} - \frac{\pi}{4} + 2n\pi \]
This simplifies to:
\[ \theta = \frac{\pi}{4} + 2n\pi \]

### Step 7: General Solution
Thus, the general solution for \( \theta \) is:
\[ \theta = \frac{\pi}{4} + 2n\pi \quad \text{for } n \in \mathbb{Z} \]