Solve ` sin 3x = cos 2x. (0 lt x lt 2pi).`

Answer

To solve the equation \( \sin 3x = \cos 2x \) for \( 0 < x < 2\pi \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \sin 3x = \cos 2x \]
We can rewrite \( \cos 2x \) using the co-function identity:
\[ \cos 2x = \sin\left(\frac{\pi}{2} - 2x\right) \]
Thus, we have:
\[ \sin 3x = \sin\left(\frac{\pi}{2} - 2x\right) \]

### Step 2: Set up the sine equation
Since \( \sin A = \sin B \), we can set up two equations:
1. \( 3x = \frac{\pi}{2} - 2x + 2n\pi \)
2. \( 3x = \pi - \left(\frac{\pi}{2} - 2x\right) + 2n\pi \)

where \( n \) is any integer.

### Step 3: Solve the first equation
From the first equation:
\[ 3x + 2x = \frac{\pi}{2} + 2n\pi \]
\[ 5x = \frac{\pi}{2} + 2n\pi \]
\[ x = \frac{\pi}{10} + \frac{2n\pi}{5} \]

### Step 4: Solve the second equation
From the second equation:
\[ 3x = \pi - \frac{\pi}{2} + 2x + 2n\pi \]
\[ 3x - 2x = \frac{\pi}{2} + 2n\pi \]
\[ x = \frac{\pi}{2} + 2n\pi \]

### Step 5: Find values of \( x \) in the interval \( 0 < x < 2\pi \)
Now we need to find valid values of \( x \) for both equations.

**For the first equation:**
\[ x = \frac{\pi}{10} + \frac{2n\pi}{5} \]
- For \( n = 0 \):
\[ x = \frac{\pi}{10} \approx 0.314 \quad (\text{valid}) \]
- For \( n = 1 \):
\[ x = \frac{\pi}{10} + \frac{2\pi}{5} = \frac{\pi}{10} + \frac{4\pi}{10} = \frac{5\pi}{10} = \frac{\pi}{2} \quad (\text{valid}) \]
- For \( n = 2 \):
\[ x = \frac{\pi}{10} + \frac{4\pi}{5} = \frac{\pi}{10} + \frac{8\pi}{10} = \frac{9\pi}{10} \quad (\text{valid}) \]
- For \( n = 3 \):
\[ x = \frac{\pi}{10} + \frac{6\pi}{5} = \frac{\pi}{10} + \frac{12\pi}{10} = \frac{13\pi}{10} \quad (\text{valid}) \]
- For \( n = 4 \):
\[ x = \frac{\pi}{10} + \frac{8\pi}{5} = \frac{\pi}{10} + \frac{16\pi}{10} = \frac{17\pi}{10} \quad (\text{valid}) \]
- For \( n = 5 \):
\[ x = \frac{\pi}{10} + \frac{10\pi}{5} = \frac{\pi}{10} + 2\pi = \frac{21\pi}{10} \quad (\text{not valid}) \]

**For the second equation:**
\[ x = \frac{\pi}{2} + 2n\pi \]
- For \( n = 0 \):
\[ x = \frac{\pi}{2} \quad (\text{valid}) \]
- For \( n = 1 \):
\[ x = \frac{\pi}{2} + 2\pi = \frac{5\pi}{2} \quad (\text{not valid}) \]

### Step 6: Compile the solutions
The valid solutions for \( x \) in the interval \( 0 < x < 2\pi \) are:
\[ x = \frac{\pi}{10}, \frac{\pi}{2}, \frac{9\pi}{10}, \frac{13\pi}{10}, \frac{17\pi}{10} \]

### Final Answer:
The solutions are:
\[ x = \frac{\pi}{10}, \frac{\pi}{2}, \frac{9\pi}{10}, \frac{13\pi}{10}, \frac{17\pi}{10} \]