If `sin 6 x= sin 4x - sin 2x , then x =`

Options

A. `n pi`
B. `(n + 1/2) pi/2`
C. `(n pi)/3 + pi/6`
D. none

Answer

To solve the equation \( \sin 6x = \sin 4x - \sin 2x \), we will follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the equation:
\[ \sin 6x + \sin 2x = \sin 4x \]

### Step 2: Using the Sine Addition Formula
We can use the sine addition formula:
\[ \sin A + \sin B = 2 \sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \]
Here, let \( A = 6x \) and \( B = 2x \):
\[ \sin 6x + \sin 2x = 2 \sin \left( \frac{6x + 2x}{2} \right) \cos \left( \frac{6x - 2x}{2} \right) = 2 \sin 4x \cos 2x \]
Thus, we can rewrite the equation as:
\[ 2 \sin 4x \cos 2x = \sin 4x \]

### Step 3: Factoring the Equation
We can factor out \( \sin 4x \):
\[ \sin 4x (2 \cos 2x - 1) = 0 \]

### Step 4: Setting Each Factor to Zero
Now we set each factor to zero:
1. \( \sin 4x = 0 \)
2. \( 2 \cos 2x - 1 = 0 \)

### Step 5: Solving \( \sin 4x = 0 \)
From \( \sin 4x = 0 \):
\[ 4x = n\pi \quad \Rightarrow \quad x = \frac{n\pi}{4} \]

### Step 6: Solving \( 2 \cos 2x - 1 = 0 \)
From \( 2 \cos 2x - 1 = 0 \):
\[ \cos 2x = \frac{1}{2} \]
The solutions for \( \cos 2x = \frac{1}{2} \) are:
\[ 2x = 2n\pi \pm \frac{\pi}{3} \quad \Rightarrow \quad x = n\pi \pm \frac{\pi}{6} \]

### Step 7: Combining Solutions
Thus, the general solutions for \( x \) are:
1. \( x = \frac{n\pi}{4} \)
2. \( x = n\pi + \frac{\pi}{6} \)
3. \( x = n\pi - \frac{\pi}{6} \)

### Final Solution
The complete solution set for \( x \) is:
\[ x = \frac{n\pi}{4}, \quad x = n\pi + \frac{\pi}{6}, \quad x = n\pi - \frac{\pi}{6} \]