Find the general solution : `sin x + sin 3x + sin 5x = 0`

Explanation

To find the general solution of the equation \( \sin x + \sin 3x + \sin 5x = 0 \), we can follow these steps:

### Step 1: Use the Sum-to-Product Identities
We can use the trigonometric identity for the sum of sines:
\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
First, we will group \( \sin 5x \) and \( \sin x \):
\[ \sin 5x + \sin x = 2 \sin\left(\frac{5x + x}{2}\right) \cos\left(\frac{5x - x}{2}\right) = 2 \sin(3x) \cos(2x) \]
Thus, we can rewrite the original equation as:
\[ 2 \sin(3x) \cos(2x) + \sin 3x = 0 \]

### Step 2: Factor the Equation
Now, we can factor out \( \sin 3x \):
\[ \sin 3x (2 \cos 2x + 1) = 0 \]

### Step 3: Set Each Factor to Zero
We have two factors to consider:
1. \( \sin 3x = 0 \)
2. \( 2 \cos 2x + 1 = 0 \)

### Step 4: Solve \( \sin 3x = 0 \)
For the first equation:
\[ \sin 3x = 0 \implies 3x = n\pi \quad (n \in \mathbb{Z}) \]
Thus,
\[ x = \frac{n\pi}{3} \]

### Step 5: Solve \( 2 \cos 2x + 1 = 0 \)
For the second equation:
\[ 2 \cos 2x + 1 = 0 \implies \cos 2x = -\frac{1}{2} \]
The general solutions for \( \cos \theta = -\frac{1}{2} \) are:
\[ 2x = \frac{2\pi}{3} + 2k\pi \quad \text{and} \quad 2x = \frac{4\pi}{3} + 2k\pi \quad (k \in \mathbb{Z}) \]
Dividing by 2 gives:
\[ x = \frac{\pi}{3} + k\pi \quad \text{and} \quad x = \frac{2\pi}{3} + k\pi \]

### Step 6: Combine the Solutions
The complete general solution is:
\[ x = \frac{n\pi}{3} \quad (n \in \mathbb{Z}) \quad \text{and} \quad x = \frac{\pi}{3} + k\pi \quad (k \in \mathbb{Z}) \quad \text{and} \quad x = \frac{2\pi}{3} + k\pi \quad (k \in \mathbb{Z}) \]

### Final General Solution
Thus, the general solution of the equation \( \sin x + \sin 3x + \sin 5x = 0 \) is:
\[ x = \frac{n\pi}{3}, \quad x = \frac{\pi}{3} + k\pi, \quad x = \frac{2\pi}{3} + k\pi \quad (n, k \in \mathbb{Z}) \]