The solution of the equation ` sin 2x + sin 4x = 2 sin 3x ` is

Options

A. `x=(npi)/3`
B. `x=npi `
C. `x=2npi `
D. None of the above

Answer

To solve the equation \( \sin 2x + \sin 4x = 2 \sin 3x \), we will follow these steps:

### Step 1: Use the sine addition formula
We know that:
\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
Here, let \( A = 4x \) and \( B = 2x \). Thus, we can rewrite the left-hand side:
\[ \sin 2x + \sin 4x = 2 \sin\left(\frac{4x + 2x}{2}\right) \cos\left(\frac{4x - 2x}{2}\right) \]
Calculating this gives:
\[ \sin 2x + \sin 4x = 2 \sin(3x) \cos(x) \]

### Step 2: Set the equation
Now we can substitute this back into the original equation:
\[ 2 \sin(3x) \cos(x) = 2 \sin(3x) \]

### Step 3: Simplify the equation
We can divide both sides by \( 2 \sin(3x) \), assuming \( \sin(3x) \neq 0 \):
\[ \cos(x) = 1 \]
If \( \sin(3x) = 0 \), we will solve that separately.

### Step 4: Solve \( \cos(x) = 1 \)
The general solution for \( \cos(x) = 1 \) is:
\[ x = 2n\pi \quad \text{where } n \text{ is an integer} \]

### Step 5: Solve \( \sin(3x) = 0 \)
The solutions for \( \sin(3x) = 0 \) are:
\[ 3x = n\pi \quad \Rightarrow \quad x = \frac{n\pi}{3} \quad \text{where } n \text{ is an integer} \]

### Final Solution
Combining both results, the complete solution set for the equation \( \sin 2x + \sin 4x = 2 \sin 3x \) is:
\[ x = 2n\pi \quad \text{and} \quad x = \frac{n\pi}{3} \quad \text{where } n \text{ is an integer} \]