The general solution of the trigonometic equation `"sin"x + "cos"x = 1` is given by

Options

A. `x = 2n pi`
B. `x= 2n pi + (pi)/(2)`
C. `x = n pi + (-1)^(n) (pi)/(4)-(pi)/(4)`
D. none of these

Answer

To solve the trigonometric equation \( \sin x + \cos x = 1 \), we will follow these steps:

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

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

### Step 3: Recognize the sine addition formula
We know that:
\[ \sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
Thus, we can rewrite the left-hand side using the sine addition formula:
\[ \sin x \cos\left(\frac{\pi}{4}\right) + \cos x \sin\left(\frac{\pi}{4}\right) = \sin\left(x + \frac{\pi}{4}\right) \]
This gives us:
\[ \sin\left(x + \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

### Step 4: Solve for \( x + \frac{\pi}{4} \)
The general solution for \( \sin \theta = k \) is:
\[ \theta = \arcsin(k) + 2n\pi \quad \text{or} \quad \theta = \pi - \arcsin(k) + 2n\pi \]
In our case, \( k = \frac{1}{\sqrt{2}} \), so:
\[ x + \frac{\pi}{4} = \frac{\pi}{4} + 2n\pi \quad \text{or} \quad x + \frac{\pi}{4} = \pi - \frac{\pi}{4} + 2n\pi \]

### Step 5: Simplify the equations
1. From \( x + \frac{\pi}{4} = \frac{\pi}{4} + 2n\pi \):
\[ x = 2n\pi \]

2. From \( x + \frac{\pi}{4} = \frac{3\pi}{4} + 2n\pi \):
\[ x = 2n\pi + \frac{3\pi}{4} - \frac{\pi}{4} = 2n\pi + \frac{\pi}{2} \]

### Step 6: Combine the solutions
Thus, the general solutions are:
\[ x = 2n\pi \quad \text{and} \quad x = 2n\pi + \frac{\pi}{2} \]

### Final General Solution
The final general solution can be expressed as:
\[ x = n\pi + \frac{\pi}{4} \quad \text{for } n \in \mathbb{Z} \]