Determine the smallest positive value of `(32x)/(pi)` which satisfy the equation `sqrt(1+sin 2x)-sqrt(2)cos 3x=0`

Answer

To solve the equation \( \sqrt{1 + \sin 2x} - \sqrt{2} \cos 3x = 0 \) and determine the smallest positive value of \( \frac{32x}{\pi} \), we can follow these steps:

### Step 1: Rearranging the Equation
Start by rearranging the equation:
\[ \sqrt{1 + \sin 2x} = \sqrt{2} \cos 3x \]

### Step 2: Squaring Both Sides
Square both sides to eliminate the square roots:
\[ 1 + \sin 2x = 2 \cos^2 3x \]

### Step 3: Using Trigonometric Identities
Recall that \( \sin 2x = 2 \sin x \cos x \) and \( \cos^2 3x = 1 - \sin^2 3x \). Substitute these into the equation:
\[ 1 + 2 \sin x \cos x = 2(1 - \sin^2 3x) \]

### Step 4: Simplifying the Equation
Rearranging gives:
\[ 1 + 2 \sin x \cos x = 2 - 2 \sin^2 3x \]
\[ 2 \sin x \cos x + 2 \sin^2 3x = 1 \]

### Step 5: Dividing by 2
Divide the entire equation by 2:
\[ \sin x \cos x + \sin^2 3x = \frac{1}{2} \]

### Step 6: Using the Double Angle Identity
Using the identity \( \sin x \cos x = \frac{1}{2} \sin 2x \):
\[ \frac{1}{2} \sin 2x + \sin^2 3x = \frac{1}{2} \]
Multiply through by 2 to eliminate the fraction:
\[ \sin 2x + 2 \sin^2 3x = 1 \]

### Step 7: Rearranging Again
Rearranging gives:
\[ \sin 2x = 1 - 2 \sin^2 3x \]

### Step 8: Using the Pythagorean Identity
Using \( 1 - 2 \sin^2 3x = \cos 6x \) (from the double angle formula):
\[ \sin 2x = \cos 6x \]

### Step 9: Setting Up the Equations
This leads to:
\[ 2x = 6x + 2n\pi \quad \text{or} \quad 2x = -6x + 2n\pi \]

### Step 10: Solving for x
From \( 2x = 6x + 2n\pi \):
\[ -4x = 2n\pi \implies x = -\frac{n\pi}{2} \]
From \( 2x = -6x + 2n\pi \):
\[ 8x = 2n\pi \implies x = \frac{n\pi}{4} \]

### Step 11: Finding the Smallest Positive Value
To find the smallest positive value of \( x \):
- For \( n = 1 \), \( x = \frac{\pi}{4} \)
- For \( n = 2 \), \( x = \frac{n\pi}{4} = \frac{2\pi}{4} = \frac{\pi}{2} \)

The smallest positive \( x \) is \( \frac{\pi}{4} \).

### Step 12: Calculating \( \frac{32x}{\pi} \)
Now calculate:
\[ \frac{32x}{\pi} = \frac{32 \cdot \frac{\pi}{4}}{\pi} = \frac{32}{4} = 8 \]

Thus, the smallest positive value of \( \frac{32x}{\pi} \) that satisfies the equation is:
\[ \boxed{8} \]