If `"sin" 2x, (1)/(2) " and cos" 2x` are in A.P., then the general values of x are given by

Options

A. `n pi, n pi + (pi)/(2), n in Z`
B. `n pi, n pi + (pi)/(4), n in Z`
C. `n pi + (pi)/(4), n in Z`
D. `n pi, n in Z`

Answer

To solve the problem where \(\sin 2x\) and \(\cos 2x\) are in arithmetic progression (A.P.), we can follow these steps:

### Step 1: Understanding the A.P. Condition
If \(\sin 2x\), \(\frac{1}{2}\), and \(\cos 2x\) are in A.P., then by the definition of A.P., we have:
\[ \frac{\sin 2x + \cos 2x}{2} = \frac{1}{2} \]

### Step 2: Simplifying the A.P. Condition
Multiplying both sides by 2 gives:
\[ \sin 2x + \cos 2x = 1 \]

### Step 3: Rearranging the Equation
We can rearrange the equation:
\[ \sin 2x = 1 - \cos 2x \]

### Step 4: Squaring Both Sides
Now, we square both sides to eliminate the sine and cosine:
\[ \sin^2 2x = (1 - \cos 2x)^2 \]
Expanding the right side:
\[ \sin^2 2x = 1 - 2\cos 2x + \cos^2 2x \]

### Step 5: Using the Pythagorean Identity
Using the identity \(\sin^2 2x + \cos^2 2x = 1\), we can substitute \(\sin^2 2x\):
\[ 1 - \cos^2 2x = 1 - 2\cos 2x + \cos^2 2x \]

### Step 6: Simplifying the Equation
Cancelling 1 from both sides:
\[ -\cos^2 2x = -2\cos 2x + \cos^2 2x \]
Rearranging gives:
\[ 2\cos^2 2x - 2\cos 2x = 0 \]

### Step 7: Factoring the Equation
Factoring out \(2\cos 2x\):
\[ 2\cos 2x(\cos 2x - 1) = 0 \]

### Step 8: Finding the Solutions
Setting each factor to zero gives us two equations:
1. \(2\cos 2x = 0\) which simplifies to \(\cos 2x = 0\)
2. \(\cos 2x - 1 = 0\) which simplifies to \(\cos 2x = 1\)

### Step 9: Solving \(\cos 2x = 0\)
The general solution for \(\cos 2x = 0\) is:
\[ 2x = \frac{\pi}{2} + n\pi \quad \Rightarrow \quad x = \frac{\pi}{4} + \frac{n\pi}{2} \]

### Step 10: Solving \(\cos 2x = 1\)
The general solution for \(\cos 2x = 1\) is:
\[ 2x = 2n\pi \quad \Rightarrow \quad x = n\pi \]

### Final Answer
Thus, the general values of \(x\) are:
\[ x = n\pi \quad \text{and} \quad x = \frac{\pi}{4} + \frac{n\pi}{2} \]