Question: Find the range of the function $ f(x) = \frac{1}{\sin^2 x + 4\cos^2 x} $ as $ x $ ranges over all real numbers such that $ \sin^2 x + \cos^2 x = 1 $.

["Title: Finding the Range of the Function $ f(x) = \frac{1}{\sin^2 x + 4\cos^2 x} $ Using Trigonometric Identities", "Meta Description:\nExplore the range of the function $ f(x) = \frac{1}{\sin^2 x + 4\cos^2 x} $ over all real $ x $. Learn how to determine the range using trigonometric identities and optimization techniques.", "---", "### Introduction", "Understanding the range of a function is essential in mathematics and applied sciences—especially in signal processing, engineering, and optimization. In this article, we analyze the function\n$$\nf(x) = \frac{1}{\sin^2 x + 4\cos^2 x}\n$$\nand determine its range as $ x $ varies over all real numbers. Given the fundamental identity $ \sin^2 x + \cos^2 x = 1 $, we simplify the expression and use algebraic techniques to find the minimum and maximum values of $ f(x) $.", "---", "### Step 1: Use the Pythagorean Identity", "Start from the identity:\n$$\n\sin^2 x + \cos^2 x = 1\n$$\nLet $ s = \sin^2 x $, then $ \cos^2 x = 1 - s $, where $ 0 \leq s \leq 1 $. Substitute into the denominator:\n$$\n\sin^2 x + 4\cos^2 x = s + 4(1 - s) = s + 4 - 4s = 4 - 3s\n$$\nThus,\n$$\nf(x) = \frac{1}{4 - 3\sin^2 x} = \frac{1}{4 - 3s}, \quad \ ext{with } 0 \leq s \leq 1\n$$", "---", "### Step 2: Analyze the Denominator Over Its Domain", "Let $ y = 4 - 3s $. Since $ 0 \leq s \leq 1 $,\n- When $ s = 0 $, $ y = 4 $\n- When $ s = 1 $, $ y = 4 - 3(1) = 1 $", "So $ y \in [1, 4] $. The function $ f(x) = \frac{1}{y} $ is now a decreasing function over $ y \in [1, 4] $, meaning the maximum value of $ f(x) $ occurs at the minimum of $ y $, and vice versa.", "- Maximum of $ f(x) $: occurs at $ y = 1 $, so $ \max f(x) = \frac{1}{1} = 1 $\n- Minimum of $ f(x) $: occurs at $ y = 4 $, so $ \min f(x) = \frac{1}{4} $", "---", "### Step 3: Confirm All Values Are Achievable", "Since $ s = \sin^2 x $ takes every value in $ [0, 1] $, and $ y = 4 - 3s $ is continuous and strictly decreasing over this interval, $ y $ spans every value in $[1, 4]$, and thus $ f(x) = \frac{1}{4 - 3s} $ takes all values from $ \frac{1}{4} $ to $ 1 $.\nTherefore, the range of $ f(x) $ is:\n$$\n\left[ \frac{1}{4}, 1 \right]\n$$", "---", "### Step 4: Final Answer", "The range of the function\n$$\nf(x) = \frac{1}{\sin^2 x + 4\cos^2 x}\n$$\nover all real $ x $ is\n$$\n\boxed{\left[ \frac{1}{4}, 1 \right]}\n$$", "---", "### Conclusion", "By leveraging the identity $ \sin^2 x + \cos^2 x = 1 $ and transforming the expression into a rational function over a continuous interval, we efficiently determined the function’s range without calculus. This approach is both elegant and practical for evaluating trigonometric functions in optimization and modeling contexts.", "---", "Keywords:\n$ f(x) = \frac{1}{\sin^2 x + 4\cos^2 x} $, range of function, trigonometric functions, $ \sin^2 x + \cos^2 x = 1 $, optimization, periodic functions, domain analysis, Fourier analysis insight, mathematical reasoning.", "For further reading: Explore how similar techniques apply to functions like $ f(x) = a\sin^2 x + b\cos^2 x $ and their reciprocals."]