If ` sin 3 alpha =4 sin alpha sin (x+alpha ) sin(x-alpha ) `, then

Options

A. `x=npi pm (pi)/(3), n in I `
B. `x=npi pm (pi)/(6), n in I `
C. `x=npi pm (pi)/(2) , n in I `
D. None of the above

Answer

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

### Step 1: Apply the formula for \( \sin 3\alpha \)
We know that:
\[ \sin 3\alpha = 3 \sin \alpha - 4 \sin^3 \alpha \]
Thus, we can rewrite the left side:
\[ 3 \sin \alpha - 4 \sin^3 \alpha = 4 \sin \alpha \sin(x + \alpha) \sin(x - \alpha) \]

### Step 2: Simplify the right side using the sine product-to-sum formula
Using the identity \( \sin A \sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)] \), we can express:
\[ \sin(x + \alpha) \sin(x - \alpha) = \frac{1}{2}[\cos((x - \alpha) - (x + \alpha)) - \cos((x - \alpha) + (x + \alpha))] = \frac{1}{2}[\cos(-2\alpha) - \cos(2x)] \]
This simplifies to:
\[ \sin(x + \alpha) \sin(x - \alpha) = \frac{1}{2}[\cos(2\alpha) - \cos(2x)] \]

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

### Step 4: Factor out \( \sin \alpha \)
Assuming \( \sin \alpha \neq 0 \), we can divide both sides by \( \sin \alpha \):
\[ 3 - 4 \sin^2 \alpha = 2[\cos(2\alpha) - \cos(2x)] \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ 2\cos(2x) = 2\cos(2\alpha) - (3 - 4 \sin^2 \alpha) \]
This simplifies to:
\[ \cos(2x) = \cos(2\alpha) - \frac{3 - 4 \sin^2 \alpha}{2} \]

### Step 6: Solve for \( x \)
To find \( x \), we can use the cosine inverse:
\[ 2x = \cos^{-1}\left(\cos(2\alpha) - \frac{3 - 4 \sin^2 \alpha}{2}\right) \]
Thus:
\[ x = \frac{1}{2} \cos^{-1}\left(\cos(2\alpha) - \frac{3 - 4 \sin^2 \alpha}{2}\right) \]

### Step 7: General solution
The general solution for \( x \) can be expressed as:
\[ x = n\pi \pm \frac{1}{2} \cos^{-1}\left(\cos(2\alpha) - \frac{3 - 4 \sin^2 \alpha}{2}\right) \]
where \( n \) is any integer.