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

Explanation

To solve the equation \( \sin x + \sin 3x + \sin 5x = 0 \), we can use the sum-to-product identities and properties of trigonometric functions. Here’s a step-by-step solution:

### Step 1: Combine the Sine Functions
We start with the equation:
\[ \sin x + \sin 3x + \sin 5x = 0 \]
We can group \( \sin x \) and \( \sin 5x \) together:
\[ \sin x + \sin 5x + \sin 3x = 0 \]

### Step 2: Use the Sum-to-Product Identity
Using the sum-to-product identities, we can combine \( \sin x \) and \( \sin 5x \):
\[ \sin x + \sin 5x = 2 \sin\left(\frac{x + 5x}{2}\right) \cos\left(\frac{5x - x}{2}\right) = 2 \sin(3x) \cos(2x) \]
Thus, we can rewrite the equation as:
\[ 2 \sin(3x) \cos(2x) + \sin(3x) = 0 \]

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

### Step 4: Set Each Factor to Zero
This gives us two equations to solve:
1. \( \sin(3x) = 0 \)
2. \( 2 \cos(2x) + 1 = 0 \)

### Step 5: Solve \( \sin(3x) = 0 \)
The sine function is zero at integer multiples of \( \pi \):
\[ 3x = n\pi \quad \Rightarrow \quad x = \frac{n\pi}{3}, \quad n \in \mathbb{Z} \]

### Step 6: Solve \( 2 \cos(2x) + 1 = 0 \)
Rearranging gives:
\[ \cos(2x) = -\frac{1}{2} \]
The cosine function is equal to \(-\frac{1}{2}\) at:
\[ 2x = \frac{2\pi}{3} + 2k\pi \quad \text{and} \quad 2x = \frac{4\pi}{3} + 2k\pi, \quad k \in \mathbb{Z} \]
Dividing by 2 gives:
\[ x = \frac{\pi}{3} + k\pi \quad \text{and} \quad x = \frac{2\pi}{3} + k\pi \]

### Step 7: Combine Solutions
Now we combine all solutions:
1. From \( \sin(3x) = 0 \): \( x = \frac{n\pi}{3} \)
2. From \( 2 \cos(2x) + 1 = 0 \):
- \( x = \frac{\pi}{3} + k\pi \)
- \( x = \frac{2\pi}{3} + k\pi \)

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