The general solution of the equation
`sin theta + cos theta = 1` is `theta` =

Options

A. `n pi + [(-1)^(n) + 1](pi//4)`
B. `2n pi + (pi//2)`
C. `2npi`
D. `n pi + [(-1)^(n) - 1](pi//4)`

Answer

To solve the equation \( \sin \theta + \cos \theta = 1 \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \sin \theta + \cos \theta = 1 \]

### Step 2: Multiply by \( \frac{1}{\sqrt{2}} \)
To simplify the equation, we can multiply both sides by \( \frac{1}{\sqrt{2}} \):
\[ \frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta = \frac{1}{\sqrt{2}} \]

### Step 3: Use the sine addition formula
Recognizing that \( \frac{1}{\sqrt{2}} = \cos \frac{\pi}{4} \) and \( \frac{1}{\sqrt{2}} = \sin \frac{\pi}{4} \), we can rewrite the left side using the sine addition formula:
\[ \sin \left( \theta + \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} \]

### Step 4: Find the general solution
The equation \( \sin x = k \) has a general solution of:
\[ x = n\pi + (-1)^n \arcsin(k) \]
In our case, \( k = \frac{1}{\sqrt{2}} \), which corresponds to \( \arcsin\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4} \). Therefore, we have:
\[ \theta + \frac{\pi}{4} = n\pi + (-1)^n \frac{\pi}{4} \]

### Step 5: Solve for \( \theta \)
Rearranging gives us:
\[ \theta = n\pi + (-1)^n \frac{\pi}{4} - \frac{\pi}{4} \]
This simplifies to:
\[ \theta = n\pi + (-1)^n \frac{\pi}{4} - \frac{\pi}{4} = n\pi + \left((-1)^n - 1\right) \frac{\pi}{4} \]

### Step 6: Final expression
Factoring out \( \frac{\pi}{4} \):
\[ \theta = n\pi + (-1)^{n-1} \frac{\pi}{4} \]
Thus, the general solution of the equation is:
\[ \theta = n\pi + (-1)^{n-1} \frac{\pi}{4} \]