If `sin x +cos x =1`, then x =

Options

A. `n pi`
B. `2n pi`
C. `2n pi + pi/2`
D. `2n pi - pi/2`

Answer

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

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

### Step 2: Square both sides
To eliminate the sine and cosine terms, we can square both sides:
\[ (\sin x + \cos x)^2 = 1^2 \]
Expanding the left side gives:
\[ \sin^2 x + 2 \sin x \cos x + \cos^2 x = 1 \]

### Step 3: Use the Pythagorean identity
Using the identity \( \sin^2 x + \cos^2 x = 1 \), we can simplify:
\[ 1 + 2 \sin x \cos x = 1 \]
Subtracting 1 from both sides results in:
\[ 2 \sin x \cos x = 0 \]

### Step 4: Factor the equation
We can factor the equation:
\[ \sin(2x) = 0 \]

### Step 5: Solve for \( x \)
The general solution for \( \sin(2x) = 0 \) is:
\[ 2x = n\pi \quad \text{where } n \text{ is an integer} \]
Dividing both sides by 2 gives:
\[ x = \frac{n\pi}{2} \]

### Step 6: Find specific solutions
From the original equation \( \sin x + \cos x = 1 \), we can check specific values:
1. For \( n = 0 \): \( x = 0 \)
2. For \( n = 1 \): \( x = \frac{\pi}{2} \)
3. For \( n = 2 \): \( x = \pi \)
4. For \( n = 3 \): \( x = \frac{3\pi}{2} \)

### Step 7: Verify solutions
We verify which of these values satisfy the original equation:
- For \( x = 0 \): \( \sin(0) + \cos(0) = 0 + 1 = 1 \) (valid)
- For \( x = \frac{\pi}{2} \): \( \sin\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right) = 1 + 0 = 1 \) (valid)
- For \( x = \pi \): \( \sin(\pi) + \cos(\pi) = 0 - 1 = -1 \) (not valid)
- For \( x = \frac{3\pi}{2} \): \( \sin\left(\frac{3\pi}{2}\right) + \cos\left(\frac{3\pi}{2}\right) = -1 + 0 = -1 \) (not valid)

### Final solutions
The valid solutions are:
\[ x = 0 + 2n\pi \quad \text{and} \quad x = \frac{\pi}{2} + 2n\pi \]