The set of values of `x` satisfying the equation `sin3alpha=4sinalphasin(x+alpha)sin(x-alpha)` is

Options

A. `npi+-pi/4,AAn in Z`
B. `npi+-pi/3,AAn in Z`
C. `npi+-pi/9,AAn in Z`
D. `npi+-pi/(12),AAn in Z`

Answer

To solve the equation \( \sin 3\alpha = 4 \sin \alpha \sin(x + \alpha) \sin(x - \alpha) \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \sin 3\alpha = 4 \sin \alpha \sin(x + \alpha) \sin(x - \alpha) \]

### Step 2: Use the product-to-sum identities
We can use the identity for the product of sine functions:
\[ \sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)] \]
Applying this to \( \sin(x + \alpha) \sin(x - \alpha) \):
\[ \sin(x + \alpha) \sin(x - \alpha) = \frac{1}{2} [\cos((x + \alpha) - (x - \alpha)) - \cos((x + \alpha) + (x - \alpha))] \]
This simplifies to:
\[ \sin(x + \alpha) \sin(x - \alpha) = \frac{1}{2} [\cos(2\alpha) - \cos(2x)] \]

### Step 3: Substitute back into the equation
Substituting this back into the original equation gives us:
\[ \sin 3\alpha = 4 \sin \alpha \cdot \frac{1}{2} [\cos(2\alpha) - \cos(2x)] \]
This simplifies to:
\[ \sin 3\alpha = 2 \sin \alpha (\cos(2\alpha) - \cos(2x)) \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ \sin 3\alpha = 2 \sin \alpha \cos(2\alpha) - 2 \sin \alpha \cos(2x) \]
We can isolate the term involving \( \cos(2x) \):
\[ 2 \sin \alpha \cos(2x) = 2 \sin \alpha \cos(2\alpha) - \sin 3\alpha \]

### Step 5: Divide by \( 2 \sin \alpha \) (assuming \( \sin \alpha \neq 0 \))
We can divide both sides by \( 2 \sin \alpha \):
\[ \cos(2x) = \cos(2\alpha) - \frac{\sin 3\alpha}{2 \sin \alpha} \]

### Step 6: Solve for \( x \)
Using the identity \( \cos A = \cos B \) gives us two cases:
1. \( 2x = 2\alpha + 2n\pi \)
2. \( 2x = -2\alpha + 2n\pi \)

From these, we can solve for \( x \):
1. \( x = \alpha + n\pi \)
2. \( x = -\alpha + n\pi \)

### Final Result
Thus, the set of values for \( x \) satisfying the equation is:
\[ x = \alpha + n\pi \quad \text{and} \quad x = -\alpha + n\pi, \quad n \in \mathbb{Z} \]