Let `f:R to R ` be the function defined by ` f(x) = (1)/(2-cos x), AA x in R.` Then, find the range of `f`.

Explanation

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

### Step 1: Set up the equation
Let \( y = f(x) = \frac{1}{2 - \cos x} \).

### Step 2: Rearranging the equation
Rearranging this equation gives us:
\[ y(2 - \cos x) = 1 \]
This simplifies to:
\[ 2y - y \cos x = 1 \]
From this, we can express \( \cos x \) in terms of \( y \):
\[ y \cos x = 2y - 1 \implies \cos x = \frac{2y - 1}{y} \]

### Step 3: Determine the range of \( \cos x \)
We know that the cosine function has a range of \([-1, 1]\). Therefore, we need to set up the inequality:
\[ -1 \leq \frac{2y - 1}{y} \leq 1 \]

### Step 4: Solve the inequalities
We will solve the two inequalities separately.

1. **First Inequality**:
\[ -1 \leq \frac{2y - 1}{y} \]
Multiplying through by \( y \) (noting that \( y > 0 \)):
\[ -y \leq 2y - 1 \implies -y - 2y \leq -1 \implies -3y \leq -1 \implies y \geq \frac{1}{3} \]

2. **Second Inequality**:
\[ \frac{2y - 1}{y} \leq 1 \]
Again, multiplying through by \( y \):
\[ 2y - 1 \leq y \implies 2y - y \leq 1 \implies y \leq 1 \]

### Step 5: Combine the results
From the two inequalities, we have:
\[ \frac{1}{3} \leq y \leq 1 \]
Thus, the range of \( f \) is:
\[ \text{Range of } f = \left[ \frac{1}{3}, 1 \right] \]

### Final Answer
The range of the function \( f(x) = \frac{1}{2 - \cos x} \) is \(\left[ \frac{1}{3}, 1 \right]\).

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