Determine the smallest positive value of `x` which satisfy the equation `sqrt(1+sin2x)-sqrt(2)cos3x=0`

Explanation

To solve the equation \( \sqrt{1 + \sin 2x} - \sqrt{2} \cos 3x = 0 \), we will follow these steps:

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

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

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

### Step 4: Rearranging the Equation
Rearranging gives:
\[ 2 \sin x \cos x + 2 \sin^2 3x - 1 = 0 \]

### Step 5: Simplifying Further
Divide the entire equation by 2:
\[ \sin x \cos x + \sin^2 3x - \frac{1}{2} = 0 \]

### Step 6: Using the Double Angle Identity
Recall that \( \sin x \cos x = \frac{1}{2} \sin 2x \):
\[ \frac{1}{2} \sin 2x + \sin^2 3x - \frac{1}{2} = 0 \]
Multiplying through by 2 gives:
\[ \sin 2x + 2 \sin^2 3x - 1 = 0 \]

### Step 7: Solving for \( \sin 2x \)
Rearranging gives:
\[ \sin 2x = 1 - 2 \sin^2 3x \]

### Step 8: Using the Identity for \( \sin 2x \)
Using the identity \( \sin 2x = 2 \sin x \cos x \), we can set up the equation:
\[ 2 \sin x \cos x = 1 - 2 \sin^2 3x \]

### Step 9: Finding Values of \( x \)
To find the smallest positive value of \( x \), we can analyze the equation further. We can set \( \sin 2x = 0 \) or \( 1 - 2 \sin^2 3x = 0 \) to find potential solutions.

1. **For \( \sin 2x = 0 \)**:
\[ 2x = n\pi \implies x = \frac{n\pi}{2} \]

2. **For \( 1 - 2 \sin^2 3x = 0 \)**:
\[ 2 \sin^2 3x = 1 \implies \sin^2 3x = \frac{1}{2} \implies \sin 3x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2} \]
This gives:
\[ 3x = \frac{\pi}{4} + k\pi \implies x = \frac{\pi}{12} + \frac{k\pi}{3} \]

### Step 10: Finding the Smallest Positive Solution
Now we need to find the smallest positive \( x \) from both sets of solutions:
- From \( x = \frac{n\pi}{2} \), the smallest positive value is \( \frac{\pi}{2} \) (for \( n=1 \)).
- From \( x = \frac{\pi}{12} + \frac{k\pi}{3} \), for \( k=0 \), we have \( \frac{\pi}{12} \) and for \( k=1 \), we have \( \frac{5\pi}{12} \).

The smallest positive value among \( \frac{\pi}{2}, \frac{\pi}{12}, \frac{5\pi}{12} \) is:
\[ \frac{\pi}{12} \]

### Final Answer
Thus, the smallest positive value of \( x \) that satisfies the equation is:
\[ \boxed{\frac{\pi}{12}} \]