Solve the equation `( sin x + cos x )^(1+ sin 2x )=2 ` , when ` -pi le x le pi ` .

Options

A. `(pi)/(2)`
B. `pi`
C. `(pi)/(4)`
D. none of these

Explanation

To solve the equation \( ( \sin x + \cos x )^{1 + \sin 2x} = 2 \) for \( -\pi \leq x \leq \pi \), we can follow these steps:

### Step 1: Rewrite the equation
The given equation is:
\[ ( \sin x + \cos x )^{1 + \sin 2x} = 2 \]
We can express \( 2 \) as \( 2^1 \). This gives us:
\[ ( \sin x + \cos x )^{1 + \sin 2x} = 2^1 \]

### Step 2: Set the bases equal
Since the bases are the same, we can equate the exponents:
\[ 1 + \sin 2x = 1 \]

### Step 3: Simplify the equation
From the equation \( 1 + \sin 2x = 1 \), we can simplify it to:
\[ \sin 2x = 0 \]

### Step 4: Solve for \( 2x \)
The general solution for \( \sin \theta = 0 \) is:
\[ \theta = n\pi \quad \text{for } n \in \mathbb{Z} \]
Thus, we have:
\[ 2x = n\pi \]

### Step 5: Solve for \( x \)
Dividing by 2 gives us:
\[ x = \frac{n\pi}{2} \]

### Step 6: Find valid solutions in the interval
Now we need to find values of \( n \) such that \( -\pi \leq x \leq \pi \):
- For \( n = -2 \): \( x = -\pi \)
- For \( n = -1 \): \( x = -\frac{\pi}{2} \)
- For \( n = 0 \): \( x = 0 \)
- For \( n = 1 \): \( x = \frac{\pi}{2} \)
- For \( n = 2 \): \( x = \pi \)

### Step 7: List all solutions
The solutions for \( x \) in the interval \( -\pi \leq x \leq \pi \) are:
\[ x = -\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi \]

### Final Answer
Thus, the solutions to the equation \( ( \sin x + \cos x )^{1 + \sin 2x} = 2 \) in the interval \( -\pi \leq x \leq \pi \) are:
\[ x = -\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi \]