To find the domain and range of the function $f(x) = \frac{1}{1 - 2\cos(x)}$, let's analyze it step by step:

Domain:

The domain of $f(x)$ is determined by the condition that the denominator cannot be zero: $1 - 2\cos(x) \neq 0$ This gives: $\cos(x) \neq \frac{1}{2}$ The cosine function equals $\frac{1}{2}$ at: $x \neq 2n\pi \pm \frac{\pi}{3}, \quad n \in \mathbb{Z}$ Thus, the domain of $f(x)$ is: $x \in \mathbb{R} \setminus \{ 2n\pi \pm \frac{\pi}{3} \, | \, n \in \mathbb{Z} \}.$

Range:

To find the range, we analyze the behavior of $f(x)$: $f(x) = \frac{1}{1 - 2\cos(x)}.$ The cosine function satisfies $-1 \leq \cos(x) \leq 1$, so: $1 - 2\cos(x) \in [-1, 3].$ However, the denominator cannot equal zero ($\cos(x) \neq \frac{1}{2}$), so: $1 - 2\cos(x) \in [-1, \frac{1}{2}) \cup (\frac{1}{2}, 3].$ The reciprocal $\frac{1}{y}$ inverts and flips the intervals, giving: $f(x) \in (-\infty, -1] \cup \left[\frac{1}{3}, \infty\right).$

Final Answer:

• Domain: $x \in \mathbb{R} \setminus \{ 2n\pi \pm \frac{\pi}{3} \, | \, n \in \mathbb{Z} \}.$
• Range: $f(x) \in (-\infty, -1] \cup \left[\frac{1}{3}, \infty\right).$

Would you like a detailed explanation of the derivation steps?

Related Questions:

1. How does the periodicity of $\cos(x)$ affect the domain?
2. Can we graph $f(x)$ to visualize the domain and range?
3. How do we handle functions with reciprocal trigonometric components?
4. What happens if $\cos(x) = 1$ or $\cos(x) = -1$?
5. What are the vertical asymptotes of $f(x)$?

Tip:

To analyze the range of functions involving trigonometric reciprocals, consider the intervals where the denominator becomes positive or negative.