Find the general solution of ` sin x + sin 5x = sin 2 x + sin 4 x `.

Answer

To solve the equation \( \sin x + \sin 5x = \sin 2x + \sin 4x \), we will follow these steps:

### Step 1: Apply the Sum-to-Product Identities
We will use the sum-to-product identities for sine, which states:
\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]

#### Left Side:
For \( \sin x + \sin 5x \):
- Let \( A = x \) and \( B = 5x \).
- Then, we have:
\[ \sin x + \sin 5x = 2 \sin\left(\frac{x + 5x}{2}\right) \cos\left(\frac{x - 5x}{2}\right) = 2 \sin(3x) \cos(2x) \]

#### Right Side:
For \( \sin 2x + \sin 4x \):
- Let \( A = 2x \) and \( B = 4x \).
- Then, we have:
\[ \sin 2x + \sin 4x = 2 \sin\left(\frac{2x + 4x}{2}\right) \cos\left(\frac{2x - 4x}{2}\right) = 2 \sin(3x) \cos(x) \]

### Step 2: Set the Two Sides Equal
Now we can set the two sides of the equation equal to each other:
\[ 2 \sin(3x) \cos(2x) = 2 \sin(3x) \cos(x) \]

### Step 3: Simplify the Equation
We can divide both sides by 2 (assuming \( \sin(3x) \neq 0 \)):
\[ \sin(3x) \cos(2x) = \sin(3x) \cos(x) \]

### Step 4: Factor Out \( \sin(3x) \)
Rearranging gives:
\[ \sin(3x) \cos(2x) - \sin(3x) \cos(x) = 0 \]
Factoring out \( \sin(3x) \):
\[ \sin(3x) (\cos(2x) - \cos(x)) = 0 \]

### Step 5: Solve Each Factor
This gives us two cases to solve:
1. \( \sin(3x) = 0 \)
2. \( \cos(2x) - \cos(x) = 0 \)

#### Case 1: Solve \( \sin(3x) = 0 \)
The general solution for \( \sin \theta = 0 \) is:
\[ 3x = n\pi \quad \Rightarrow \quad x = \frac{n\pi}{3}, \quad n \in \mathbb{Z} \]

#### Case 2: Solve \( \cos(2x) = \cos(x) \)
Using the identity \( \cos A - \cos B = 0 \):
\[ \cos(2x) = \cos(x) \quad \Rightarrow \quad 2x = 2k\pi \pm x \quad (k \in \mathbb{Z}) \]
This gives us two equations:
1. \( 2x = 2k\pi + x \) \( \Rightarrow x = 2k\pi \)
2. \( 2x = 2k\pi - x \) \( \Rightarrow 3x = 2k\pi \) \( \Rightarrow x = \frac{2k\pi}{3} \)

### Step 6: Combine Solutions
Combining all solutions:
1. From \( \sin(3x) = 0 \): \( x = \frac{n\pi}{3} \)
2. From \( \cos(2x) = \cos(x) \): \( x = 2k\pi \) and \( x = \frac{2k\pi}{3} \)

### Final General Solution
Thus, the general solutions are:
\[ x = \frac{n\pi}{3}, \quad n \in \mathbb{Z} \]
\[ x = 2k\pi, \quad k \in \mathbb{Z} \]
\[ x = \frac{2k\pi}{3}, \quad k \in \mathbb{Z} \]