Question: Find the range of $ f(x) = \frac{2\sin x + 1}{\sin x - 3} $ for $ x $ where the denominator is nonzero.

["Title: Find the Range of $ f(x) = \frac{2\sin x + 1}{\sin x - 3} $ — A Complete Guide", "Meta Description:\nDiscover the range of the function $ f(x) = \frac{2\sin x + 1}{\sin x - 3} $ where the denominator is nonzero. Learn step-by-step how to determine all possible output values using algebraic manipulation and domain analysis.", "---", "## Introduction", "Understanding the range of a function is essential in calculus and algebra, especially when dealing with trigonometric expressions. One such function is:", "[\nf(x) = \frac{2\sin x + 1}{\sin x - 3}\n]", "Since $ \sin x $ is bounded between $-1$ and $1$, but the denominator must never be zero, we must carefully analyze the behavior of $ f(x) $ over its valid domain.", "This article explains how to find the full range of $ f(x) $ when $ \sin x - 3 <br/>\ne 0 $, which is always true here since $ \sin x \in [-1, 1] $ and thus cannot equal 3.", "---", "## Domain and Key Observations", "First, note that the denominator $ \sin x - 3 $ is never zero for $ x \in \mathbb{R} $, because $ \sin x \leq 1 $, so $ \sin x - 3 \leq -2 $. Therefore, the function is defined for all real numbers, and no values of $ x $ need to be excluded.", "Our goal is to find all real values $ y $ such that:", "[\ny = \frac{2\sin x + 1}{\sin x - 3}\n]", "can be satisfied for some $ \sin x \in [-1, 1] $.", "---", "## Step 1: Let $ s = \sin x $, so $ s \in [-1, 1] $", "Substitute $ s $ for $ \sin x $:", "[\ny = \frac{2s + 1}{s - 3}\n]", "We now seek all possible $ y $ such that there exists some $ s \in [-1, 1] $ satisfying this equation.", "---", "## Step 2: Solve for $ s $ in terms of $ y $", "Rewriting:", "[\ny(s - 3) = 2s + 1\n]", "[\nys - 3y = 2s + 1\n]", "Bring all terms to one side:", "[\nys - 2s = 3y + 1\n]", "[\ns(y - 2) = 3y + 1\n]", "Assuming $ y - 2 <br/>\ne 0 $, solve for $ s $:", "[\ns = \frac{3y + 1}{y - 2}\n]", "---", "## Step 3: Ensure $ s \in [-1, 1] $", "Since $ s = \sin x $, we require:", "[\n-1 \leq \frac{3y + 1}{y - 2} \leq 1\n]", "We solve this compound inequality in two parts.", "---", "### Part A: $ \frac{3y + 1}{y - 2} \geq -1 $", "[\n\frac{3y + 1}{y - 2} + 1 \geq 0\n]", "[\n\frac{3y + 1 + (y - 2)}{y - 2} \geq 0\n]", "[\n\frac{4y - 1}{y - 2} \geq 0\n]", "### Part B: $ \frac{3y + 1}{y - 2} \leq 1 $", "[\n\frac{3y + 1}{y - 2} - 1 \leq 0\n]", "[\n\frac{3y + 1 - (y - 2)}{y - 2} \leq 0\n]", "[\n\frac{2y + 3}{y - 2} \leq 0\n]", "---", "## Step 4: Solve the Compound Inequality", "We solve:", "1. $ \frac{4y - 1}{y - 2} \geq 0 $\n2. $ \frac{2y + 3}{y - 2} \leq 0 $", "Let’s analyze each using sign charts, considering critical points at $ y = -3/2 $, $ y = 1/4 $, and $ y = 2 $ (note: $ y <br/>\ne 2 $).", "### Critical points: $ y = -1.5, 0.25, 2 $", "We analyze intervals:", "---", "### Interval 1: $ y < -1.5 $\nPick $ y = -2 $", "- $ \frac{4(-2) - 1}{-2 - 2} = \frac{-9}{-4} = 2.25 > 0 $ → satisfies (1)\n- $ \frac{2(-2) + 3}{-2 - 2} = \frac{-1}{-4} = 0.25 > 0 $ → fails (2)", "Fails", "---", "### Interval 2: $ -1.5 \leq y < 2 $\nPick $ y = 0 $", "- $ \frac{-1}{-2} = 0.5 \geq 0 $ → (1) satisfied\n- $ \frac{3}{-2} = -1.5 \leq 0 $ → (2) satisfied", "Check endpoints:", "- $ y = -1.5 $:\n $ s = \frac{3(-1.5)+1}{-1.5 - 2} = \frac{-4.5 + 1}{-3.5} = \frac{-3.5}{-3.5} = 1 \in [-1,1] $ → OK", "- $ y \ o 2^- $: denominator $ \ o 0^- $, numerator $ \ o 7 $, so $ s \ o -\infty $ → outside domain", "So entire interval $ [-1.5, 2) $ satisfies?", "But wait — at $ y = 2 $, undefined → excluded.", "Now test upper bound: say $ y = 1 $:", "[\ns = \frac{3(1)+1}{1 - 2} = \frac{4}{-1} = -4 <br/>\notin [-1,1]\n]", "Wait — contradiction? But earlier we had $ s = \frac{3y+1}{y-2} $, and at $ y = 1 $, $ s = -4 $ — invalid.", "Ah! Important catch: although $ s $ is defined algebraically, we must ensure $ s \in [-1, 1] $. So even if inequality holds, the value may lie outside the physical domain.", "So return: we must intersect the solution of the inequalities with $ s = \frac{3y+1}{y-2} \in [-1,1] $.", "Let’s instead plot or test boundaries of $ s \in [-1,1] $ and compute corresponding $ y $, or solve for $ y $ satisfying $ \frac{3y+1}{y-2} \in [-1,1] $.", "Alternatively, consider $ y = \frac{3y+1}{y - 2} $, and define:", "Let $ g(y) = \frac{3y+1}{y-2} $. We want $ g(y) \in [-1,1] $", "We already have:", "- $ \frac{4y - 1}{y - 2} \geq 0 $ → $ y \in (-\infty, -1/4] \cup (2, \infty) $ but no — correction:", "Wait — earlier sign analysis:", "$ \frac{4y - 1}{y - 2} \geq 0 $: numerator zero at $ y = 1/4 $, denominator zero at $ y = 2 $.", "Sign chart:", "- $ y < 1/4 $: $ 4y-1 < 0 $, $ y-2 < 0 $ → positive → $ \geq 0 $\n- $ 1/4 < y < 2 $: $ 4y-1 > 0 $, $ y-2 < 0 $ → negative\n- $ y > 2 $: both positive → positive", "So $ \frac{4y-1}{y-2} \geq 0 $ when $ y \leq 1/4 $ or $ y > 2 $ — but $ y > 2 $ gives $ s > 1 $, invalid.", "So only valid for $ y \leq 1/4 $", "Now second inequality: $ \frac{2y + 3}{y - 2} \leq 0 $", "Numerator zero at $ y = -3/2 $, denominator zero at $ y = 2 $", "Sign:", "- $ y < -1.5 $: numerator negative, denominator negative → positive\n- $ -1.5 < y < 2 $: numerator positive, denominator negative → negative → satisfies\n- $ y > 2 $: both positive → positive", "So $ \leq 0 $ when $ y \in [-1.5, 2) $", "---", "### Intersect $ y \leq 1/4 $ and $ y \in [-1.5, 2) $:\nIntersection: $ y \in [-1.5, 0.25] $", "Also, check endpoints:", "- $ y = -1.5 $:\n $ s = \frac{3(-1.5) + 1}{-1.5 - 2} = \frac{-4.5 + 1}{-3.5} = \frac{-3.5}{-3.5} = 1 \in [-1,1] $ — valid", "- $ y = 0.25 $:\n $ s = \frac{3(0.25)+1}{0.25 - 2} = \frac{0.75 + 1}{-1.75} = \frac{1.75}{-1.75} = -1 \in [-1,1] $ — valid", "Thus, $ y \in [-1.5, 0.25] $ satisfies both inequalities and maps to $ \sin x \in [-1,1] $", "---", "## Step 5: Check Boundary Behavior and Maximum/Minimum", "Since $ f(x) $ is continuous on $ \mathbb{R} $, and $ \sin x $ traces $ [-1,1] $ continuously, $ f(x) $ traces all values in its range over $ [-1,1] $", "We now determine if $ f(x) $ attains its full range over $ s \in [-1,1] $", "Since $ s = \frac{3y + 1}{y - 2} $ is continuous on $ [-1,1] $ (denominator never zero), and we’ve found that $ y \in [-1.5, 0.25] $ corresponds exactly to valid $ s \in [-1,1] $, the range is precisely $ [-1.5, 0.25] $", "---", "## Step 6: Confirm Extremes", "Let’s compute $ f(x) $ at endpoints to confirm:", "- When $ \sin x = -1 $:\n $ f(x) = \frac{2(-1) + 1}{-1 - 3} = \frac{-2 + 1}{-4} = \frac{-1}{-4} = 0.25 $", "- When $ \sin x = 1 $:\n $ f(x) = \frac{2(1) + 1}{1 - 3} = \frac{3}{-2} = -1.5 $", "So the function achieves the endpoints.", "---", "## Final Conclusion", "The range of $ f(x) = \frac{2\sin x + 1}{\sin x - 3} $ for all real $ x $ (where defined, always true) is:", "[\n\boxed{[-1.5,\ 0.25]}\n]", "or equivalently:", "[\n\boxed{\left[-\frac{3}{2},\ \frac{1}{4}\right]}\n]", "This result is obtained by color-coding the algebra with domain awareness, inequality analysis, and verification over trigonometric bounds.", "---", "Keywords: $ f(x) = \frac{2\sin x + 1}{\sin x - 3} $ range, find range, trigonometric function analysis, domain restriction, solving rational functions, sine substitution", "Related Topics: range of rational functions, trigonometric identities, solving for function range, domain analysis, limits and continuity in trig functions", "---", "> Pro Tip: When analyzing functions involving trigonometric expressions in the denominator, always verify the domain first, then map the valid outputs by solving inequalities on the resulting variable substitution. This prevents extraneous values and ensures mathematical rigor."]