If `"sin "3theta = 4"sin" theta("sin"^(2) x-"sin"^(2)theta), theta ne npi, n in Z`. Then, the set of values of x, is

Options

A. `{n pi +- (pi)/(3): n in Z}`
B. `{n pi +-(2pi)/(3): n in Z}`
C. `{n pi +- (pi)/(2): n in Z}`
D. `{n pi +- (pi)/(4): n in Z}`

Answer

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

### Step 1: Use the identity for \( \sin 3\theta \)
We know that:
\[ \sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta \]
Substituting this into the equation gives:
\[ 3 \sin \theta - 4 \sin^3 \theta = 4 \sin \theta (\sin^2 x - \sin^2 \theta) \]

### Step 2: Expand the right-hand side
Expanding the right-hand side:
\[ 3 \sin \theta - 4 \sin^3 \theta = 4 \sin \theta \sin^2 x - 4 \sin \theta \sin^2 \theta \]

### Step 3: Rearranging the equation
Now we can rearrange the equation:
\[ 3 \sin \theta - 4 \sin \theta \sin^2 \theta = 4 \sin \theta \sin^2 x \]
Factoring out \( \sin \theta \) (noting that \( \sin \theta \neq 0 \)):
\[ \sin \theta (3 - 4 \sin^2 \theta) = 4 \sin \theta \sin^2 x \]

### Step 4: Cancel \( \sin \theta \)
Since \( \sin \theta \neq 0 \), we can divide both sides by \( \sin \theta \):
\[ 3 - 4 \sin^2 \theta = 4 \sin^2 x \]

### Step 5: Isolate \( \sin^2 x \)
Rearranging gives:
\[ 4 \sin^2 x = 3 - 4 \sin^2 \theta \]
Thus,
\[ \sin^2 x = \frac{3 - 4 \sin^2 \theta}{4} \]

### Step 6: Express \( \sin^2 x \) in terms of a known sine value
We can rewrite \( \sin^2 x \):
\[ \sin^2 x = \frac{3}{4} - \sin^2 \theta \]
Recognizing that \( \frac{3}{4} = \left(\frac{\sqrt{3}}{2}\right)^2 \), we can express it as:
\[ \sin^2 x = \left(\frac{\sqrt{3}}{2}\right)^2 - \sin^2 \theta \]

### Step 7: General solution for \( x \)
Since \( \sin^2 x = \sin^2 \alpha \) where \( \alpha = \frac{\pi}{3} \), the general solution for \( x \) is:
\[ x = n\pi \pm \frac{\pi}{3}, \quad n \in \mathbb{Z} \]

### Final Answer
Thus, the set of values of \( x \) is:
\[ x = n\pi + \frac{\pi}{3} \quad \text{or} \quad x = n\pi - \frac{\pi}{3}, \quad n \in \mathbb{Z} \]