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

Answer

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

### Step 1: Determine the Domain
The function \( f(x) \) is defined as long as the denominator is not equal to zero. Therefore, we need to find when \( 2 - \cos(3x) \neq 0 \).

1. Set the denominator equal to zero:
\[ 2 - \cos(3x) = 0 \]
This simplifies to:
\[ \cos(3x) = 2 \]

2. Since the range of the cosine function is \([-1, 1]\), \( \cos(3x) \) can never equal 2. Hence, there are no restrictions on \( x \).

Thus, the domain of \( f(x) \) is:
\[ \text{Domain: } x \in \mathbb{R} \]

### Step 2: Determine the Range
Next, we will find the range of the function. We start by expressing \( f(x) \) in terms of \( y \):
\[ y = \frac{1}{2 - \cos(3x)} \]

1. Rearranging gives:
\[ 2 - \cos(3x) = \frac{1}{y} \]
Thus,
\[ \cos(3x) = 2 - \frac{1}{y} \]

2. Since \( \cos(3x) \) must also lie within the range \([-1, 1]\), we set up the inequalities:
\[ -1 \leq 2 - \frac{1}{y} \leq 1 \]

3. Solving the left inequality:
\[ -1 \leq 2 - \frac{1}{y} \]
Rearranging gives:
\[ -3 \leq -\frac{1}{y} \implies 3 \geq \frac{1}{y} \implies y \geq \frac{1}{3} \]

4. Solving the right inequality:
\[ 2 - \frac{1}{y} \leq 1 \]
Rearranging gives:
\[ -\frac{1}{y} \leq -1 \implies \frac{1}{y} \geq 1 \implies y \leq 1 \]

Combining these results, we find:
\[ \frac{1}{3} \leq y \leq 1 \]

Thus, the range of \( f(x) \) is:
\[ \text{Range: } \left[\frac{1}{3}, 1\right] \]

### Final Answer
- **Domain:** \( x \in \mathbb{R} \)
- **Range:** \( \left[\frac{1}{3}, 1\right] \)