Number of solutions of the equation `sin x + cos x-2sqrt(2) sin x cos x=0` for `x in [0, pi]` is

Options

A. 3
B. 0
C. 1
D. 2

Explanation

To find the number of solutions for the equation \( \sin x + \cos x - 2\sqrt{2} \sin x \cos x = 0 \) within the interval \( x \in [0, \pi] \), we can follow these steps:

### Step 1: Rewrite the Equation
We start with the given equation:
\[ \sin x + \cos x - 2\sqrt{2} \sin x \cos x = 0 \]
Rearranging gives us:
\[ \sin x + \cos x = 2\sqrt{2} \sin x \cos x \]

### Step 2: Use the Identity for \( \sin x + \cos x \)
Recall the identity:
\[ \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \]
Thus, we can rewrite the equation as:
\[ \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) = 2\sqrt{2} \sin x \cos x \]
Using the double angle identity \( \sin 2x = 2 \sin x \cos x \), we have:
\[ \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) = \sqrt{2} \sin 2x \]

### Step 3: Simplify the Equation
Dividing both sides by \( \sqrt{2} \) (since \( \sqrt{2} \neq 0 \)):
\[ \sin\left(x + \frac{\pi}{4}\right) = \sin 2x \]

### Step 4: Set Up the General Solutions
The equation \( \sin A = \sin B \) gives us the general solutions:
\[ A = B + 2k\pi \quad \text{or} \quad A = \pi - B + 2k\pi \quad (k \in \mathbb{Z}) \]
Applying this to our equation:
1. \( x + \frac{\pi}{4} = 2x + 2k\pi \)
2. \( x + \frac{\pi}{4} = \pi - 2x + 2k\pi \)

### Step 5: Solve Each Case
#### Case 1:
From \( x + \frac{\pi}{4} = 2x + 2k\pi \):
\[ \frac{\pi}{4} = x + 2k\pi \implies x = \frac{\pi}{4} - 2k\pi \]
For \( k = 0 \):
\[ x = \frac{\pi}{4} \quad (\text{valid since } 0 \leq \frac{\pi}{4} \leq \pi) \]

#### Case 2:
From \( x + \frac{\pi}{4} = \pi - 2x + 2k\pi \):
\[ 3x = \pi - \frac{\pi}{4} + 2k\pi \implies 3x = \frac{3\pi}{4} + 2k\pi \implies x = \frac{\pi}{4} + \frac{2k\pi}{3} \]
For \( k = 0 \):
\[ x = \frac{\pi}{4} \quad (\text{valid}) \]
For \( k = 1 \):
\[ x = \frac{\pi}{4} + \frac{2\pi}{3} = \frac{3\pi}{12} + \frac{8\pi}{12} = \frac{11\pi}{12} \quad (\text{valid since } 0 \leq \frac{11\pi}{12} \leq \pi) \]

### Step 6: Count the Solutions
The valid solutions we found in the interval \( [0, \pi] \) are:
1. \( x = \frac{\pi}{4} \)
2. \( x = \frac{11\pi}{12} \)

Thus, the total number of solutions is **2**.

### Final Answer
The number of solutions of the equation \( \sin x + \cos x - 2\sqrt{2} \sin x \cos x = 0 \) for \( x \in [0, \pi] \) is **2**.