Find the domain and range of the function, `f(x)= (1)/(3-sin 2x)`.

Answer

To find the domain and range of the function \( f(x) = \frac{1}{3 - \sin 2x} \), we will follow these steps:

### Step 1: Determine the Domain
The domain of a function consists of all the possible input values (x-values) for which the function is defined. Since \( f(x) \) is a rational function, we need to ensure that the denominator does not equal zero.

1. Set the denominator equal to zero:
\[ 3 - \sin 2x = 0 \]

2. Solve for \( \sin 2x \):
\[ \sin 2x = 3 \]

3. Analyze the equation. The sine function, \( \sin \theta \), has a range of values between -1 and 1 for all real numbers \( \theta \). Since 3 is outside this range, there are no values of \( x \) for which \( \sin 2x = 3 \).

Thus, the function is defined for all real numbers \( x \).

**Domain:**
\[ \text{Domain} = \mathbb{R} \quad \text{(all real numbers)} \]

### Step 2: Determine the Range
The range of a function is the set of all possible output values (f(x) values). To find the range of \( f(x) \), we will analyze how \( f(x) \) behaves based on the values of \( \sin 2x \).

1. The sine function oscillates between -1 and 1:
\[ -1 \leq \sin 2x \leq 1 \]

2. Substitute the minimum and maximum values of \( \sin 2x \) into \( f(x) \):
- When \( \sin 2x = -1 \):
\[ f(x) = \frac{1}{3 - (-1)} = \frac{1}{3 + 1} = \frac{1}{4} \]

- When \( \sin 2x = 1 \):
\[ f(x) = \frac{1}{3 - 1} = \frac{1}{2} \]

3. Therefore, as \( \sin 2x \) varies from -1 to 1, \( f(x) \) will take values from \( \frac{1}{4} \) to \( \frac{1}{2} \).

**Range:**
\[ \text{Range} = \left[ \frac{1}{4}, \frac{1}{2} \right] \quad \text{(inclusive)} \]

### Summary
- **Domain:** \( \mathbb{R} \) (all real numbers)
- **Range:** \( \left[ \frac{1}{4}, \frac{1}{2} \right] \)