Find the general solution of the following equations
`sin 7x = sin 3x`

Answer

To find the general solution of the equation \( \sin 7x = \sin 3x \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \sin 7x - \sin 3x = 0 \]
This can be rewritten using the sine subtraction formula.

### Step 2: Apply the sine subtraction formula
Using the identity \( \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \), we have:
\[ \sin 7x - \sin 3x = 2 \cos\left(\frac{7x + 3x}{2}\right) \sin\left(\frac{7x - 3x}{2}\right) \]
This simplifies to:
\[ 2 \cos(5x) \sin(2x) = 0 \]

### Step 3: Set each factor to zero
For the product to be zero, either factor must be zero:
1. \( \cos(5x) = 0 \)
2. \( \sin(2x) = 0 \)

### Step 4: Solve \( \cos(5x) = 0 \)
The cosine function is zero at odd multiples of \( \frac{\pi}{2} \):
\[ 5x = \left(2n + 1\right) \frac{\pi}{2}, \quad n \in \mathbb{Z} \]
Dividing both sides by 5 gives:
\[ x = \frac{(2n + 1) \pi}{10}, \quad n \in \mathbb{Z} \]

### Step 5: Solve \( \sin(2x) = 0 \)
The sine function is zero at integer multiples of \( \pi \):
\[ 2x = n\pi, \quad n \in \mathbb{Z} \]
Dividing both sides by 2 gives:
\[ x = \frac{n\pi}{2}, \quad n \in \mathbb{Z} \]

### Step 6: Combine the solutions
The general solutions to the equation \( \sin 7x = \sin 3x \) are:
\[ x = \frac{(2n + 1) \pi}{10}, \quad n \in \mathbb{Z} \]
and
\[ x = \frac{n\pi}{2}, \quad n \in \mathbb{Z} \]

### Final Answer
The general solution is:
\[ x = \frac{(2n + 1) \pi}{10} \quad \text{and} \quad x = \frac{n\pi}{2}, \quad n \in \mathbb{Z} \]