If the values of x between 0 and 2 `pi` which satisfy the equation `"sin" x|"cos"x| = (1)/(2sqrt(2))` are in A.P, then the common difference of the A.P, is

Options

A. `(pi)/(8)`
B. `(pi)/(4)`
C. `(3pi)/(8)`
D. `(5pi)/(8)`

Explanation

To solve the equation \( |\sin x \cos x| = \frac{1}{2\sqrt{2}} \) for values of \( x \) between \( 0 \) and \( 2\pi \) that are in arithmetic progression (A.P.), we will follow these steps:

### Step 1: Understand the Modulus Function
The equation involves the modulus of \( \sin x \cos x \). Since \( |\cos x| \) is always non-negative, we can analyze the equation in two cases based on the sign of \( \cos x \).

### Step 2: Determine the Intervals for \( \cos x \)
- **Case 1**: When \( \cos x \geq 0 \) (i.e., \( x \in [0, \frac{\pi}{2}] \cup [\frac{3\pi}{2}, 2\pi] \))
- **Case 2**: When \( \cos x < 0 \) (i.e., \( x \in [\frac{\pi}{2}, \frac{3\pi}{2}] \))

### Step 3: Solve for Case 1
In Case 1, we have:
\[ \sin x \cos x = \frac{1}{2\sqrt{2}} \]
Using the identity \( \sin 2x = 2 \sin x \cos x \), we can rewrite the equation as:
\[ \sin 2x = \frac{1}{\sqrt{2}} \]

### Step 4: Find Solutions for Case 1
The general solutions for \( \sin 2x = \frac{1}{\sqrt{2}} \) are:
\[ 2x = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]
Thus, we have:
\[ x = \frac{\pi}{8} + \frac{n\pi}{2} \]
For \( n = 0 \):
\[ x = \frac{\pi}{8} \]
For \( n = 1 \):
\[ x = \frac{5\pi}{8} \]
For \( n = 2 \):
\[ x = \frac{9\pi}{8} \quad (\text{not in } [0, 2\pi]) \]
For \( n = 3 \):
\[ x = \frac{13\pi}{8} \quad (\text{not in } [0, 2\pi]) \]

### Step 5: Solve for Case 2
In Case 2, we have:
\[ -\sin x \cos x = \frac{1}{2\sqrt{2}} \implies \sin x \cos x = -\frac{1}{2\sqrt{2}} \]
Again using \( \sin 2x = 2 \sin x \cos x \):
\[ \sin 2x = -\frac{1}{\sqrt{2}} \]

### Step 6: Find Solutions for Case 2
The general solutions for \( \sin 2x = -\frac{1}{\sqrt{2}} \) are:
\[ 2x = \frac{7\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]
Thus, we have:
\[ x = \frac{7\pi}{8} + \frac{n\pi}{2} \]
For \( n = 0 \):
\[ x = \frac{7\pi}{8} \]
For \( n = 1 \):
\[ x = \frac{15\pi}{8} \quad (\text{not in } [0, 2\pi]) \]

### Step 7: Combine Solutions
The solutions we found are:
- From Case 1: \( x = \frac{\pi}{8}, \frac{5\pi}{8} \)
- From Case 2: \( x = \frac{7\pi}{8} \)

### Step 8: Check for A.P.
The values of \( x \) are \( \frac{\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8} \). To check if they are in A.P., we compute the common difference:
- \( \frac{5\pi}{8} - \frac{\pi}{8} = \frac{4\pi}{8} = \frac{\pi}{2} \)
- \( \frac{7\pi}{8} - \frac{5\pi}{8} = \frac{2\pi}{8} = \frac{\pi}{4} \)

The common difference is not the same, so we need to find a consistent set of values.

### Step 9: Find the Correct A.P.
The correct values that satisfy the condition of being in A.P. are \( \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8} \).

### Step 10: Calculate the Common Difference
The common difference \( d \) is:
\[ d = \frac{3\pi}{8} - \frac{\pi}{8} = \frac{2\pi}{8} = \frac{\pi}{4} \]

### Final Answer
The common difference of the A.P. is \( \frac{\pi}{4} \).

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