If `f(x)=sin(3x)/sin x,x!=npi`, then the range of values of f(x) for real values of x is

Answer

To find the range of the function \( f(x) = \frac{\sin(3x)}{\sin x} \) for \( x \neq n\pi \), we can follow these steps:

### Step 1: Rewrite the function
Let \( f(x) = y \), so we have:
\[ y = \frac{\sin(3x)}{\sin x} \]

### Step 2: Use the triple angle formula
Using the identity for sine of triple angle, we know:
\[ \sin(3x) = 3\sin x - 4\sin^3 x \]
Substituting this into our equation gives:
\[ y = \frac{3\sin x - 4\sin^3 x}{\sin x} \]

### Step 3: Simplify the equation
We can simplify the expression:
\[ y = 3 - 4\sin^2 x \]

### Step 4: Express \( \sin^2 x \) in terms of \( y \)
Rearranging the equation, we get:
\[ 4\sin^2 x = 3 - y \]
Thus,
\[ \sin^2 x = \frac{3 - y}{4} \]

### Step 5: Determine the range of \( \sin^2 x \)
Since \( \sin^2 x \) must be in the range \( [0, 1] \), we set up the inequalities:
\[ 0 \leq \frac{3 - y}{4} \leq 1 \]

### Step 6: Solve the inequalities
1. From \( 0 \leq \frac{3 - y}{4} \):
\[ 3 - y \geq 0 \implies y \leq 3 \]

2. From \( \frac{3 - y}{4} \leq 1 \):
\[ 3 - y \leq 4 \implies y \geq -1 \]

### Step 7: Combine the results
Combining both results, we have:
\[ -1 \leq y \leq 3 \]

### Conclusion
Since \( y = f(x) \), the range of \( f(x) \) is:
\[ [-1, 3] \]