Question:
The range of the function \( f(x) = \frac{1}{2 - \cos 3x} \) is
The range of the function \( f(x) = \frac{1}{2 - \cos 3x} \) is
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When taking reciprocals, always reverse inequality signs if all terms are positive.
Updated On: Jul 5, 2026
- \( (-2,\infty) \)
- \( [-2,3] \)
- \( \left(\frac{1}{3},2\right) \)
- \( \left(\frac{1}{2},1\right) \)
- \( \left[\frac{1}{3},1\right] \)
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Solution and Explanation
Concept:
The cosine function has range:
\[
-1 \leq \cos 3x \leq 1
\]
We transform this to find the range of the function.
Step 1: Apply cosine bounds
\[ -1 \leq \cos 3x \leq 1 \]
Step 2: Transform denominator
\[ 2 - \cos 3x \] When \( \cos 3x = 1 \): \[ 2 - 1 = 1 \] When \( \cos 3x = -1 \): \[ 2 - (-1) = 3 \] So, \[ 1 \leq 2 - \cos 3x \leq 3 \]
Step 3: Take reciprocal
\[ f(x) = \frac{1}{2 - \cos 3x} \] Taking reciprocal reverses inequality: \[ \frac{1}{3} \leq f(x) \leq 1 \]
Step 4: Check endpoints
Both values are attainable since cosine reaches both -1 and 1.
Step 5: Final answer
\[ \boxed{\left[\frac{1}{3},1\right]} \]
Step 1: Apply cosine bounds
\[ -1 \leq \cos 3x \leq 1 \]
Step 2: Transform denominator
\[ 2 - \cos 3x \] When \( \cos 3x = 1 \): \[ 2 - 1 = 1 \] When \( \cos 3x = -1 \): \[ 2 - (-1) = 3 \] So, \[ 1 \leq 2 - \cos 3x \leq 3 \]
Step 3: Take reciprocal
\[ f(x) = \frac{1}{2 - \cos 3x} \] Taking reciprocal reverses inequality: \[ \frac{1}{3} \leq f(x) \leq 1 \]
Step 4: Check endpoints
Both values are attainable since cosine reaches both -1 and 1.
Step 5: Final answer
\[ \boxed{\left[\frac{1}{3},1\right]} \]
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