Solution: Let $ y = \frac{2\sin x + 1}{\sin x - 3} $. Solve for $ \sin x $: $ y(\sin x - 3) = 2\sin x + 1 \Rightarrow y\sin x - 3y = 2\sin x + 1 \Rightarrow \sin x(y - 2) = 3y + 1 $. Thus, $ \sin x = \frac{3y + 1}{y - 2} $. Since $ |\sin x| \leq 1 $, solve $ \left| \frac{3y + 1}{y - 2} \right| \leq 1 $. This leads to $ -1 \leq \frac{3y + 1}{y - 2} \leq 1 $. Sol

["Solving for $ \sin x $: A Step-by-Step Guide to the Formula $ \sin x = \frac{3y + 1}{y - 2} $", "In trigonometry, manipulating equations involving $ \sin x $ is essential for solving problems with applications in calculus, physics, and engineering. One powerful transformation involves expressing $ \sin x $ in terms of a new variable $ y = \frac{2\sin x + 1}{\sin x - 3} $. Understanding how to isolate $ \sin x $ and determine its valid domain is key.", "This article guides you through solving for $ \sin x $ from the equation:", "$$\ny = \frac{2\sin x + 1}{\sin x - 3}\n$$", "### Step 1: Rearranging the Equation", "Start by eliminating the denominator:", "$$\ny(\sin x - 3) = 2\sin x + 1\n$$", "Distribute $ y $:", "$$\ny \sin x - 3y = 2\sin x + 1\n$$", "Bring all terms involving $ \sin x $ to one side:", "$$\ny \sin x - 2\sin x = 3y + 1\n$$", "Factor $ \sin x $:", "$$\n\sin x (y - 2) = 3y + 1\n$$", "Solve for $ \sin x $:", "$$\n\sin x = \frac{3y + 1}{y - 2}\n$$", "This expression shows $ \sin x $ as a rational function of $ y $, useful for analyzing its range.", "---", "### Step 2: Applying the Domain of Sine", "Since $ \sin x $ must satisfy $ |\sin x| \leq 1 $, we impose the inequality:", "$$\n\left| \frac{3y + 1}{y - 2} \right| \leq 1\n$$", "This equivalent to:", "$$\n-1 \leq \frac{3y + 1}{y - 2} \leq 1\n$$", "We solve this compound inequality step by step.", "---", "### Step 3: Solve the Compound Inequality", "#### Part 1: $ \frac{3y + 1}{y - 2} \leq 1 $", "Subtract 1 from both sides:", "$$\n\frac{3y + 1}{y - 2} - 1 \leq 0 \quad \Rightarrow \quad \frac{3y + 1 - (y - 2)}{y - 2} \leq 0\n$$", "Simplify numerator:", "$$\n\frac{3y + 1 - y + 2}{y - 2} = \frac{2y + 3}{y - 2} \leq 0\n$$", "#### Part 2: $ \frac{3y + 1}{y - 2} \geq -1 $", "Add 1 to both sides:", "$$\n\frac{3y + 1}{y - 2} + 1 \geq 0 \quad \Rightarrow \quad \frac{3y + 1 + (y - 2)}{y - 2} \geq 0\n$$", "Simplify numerator:", "$$\n\frac{4y - 1}{y - 2} \geq 0\n$$", "---", "### Step 4: Combine the Inequalities", "We now solve the system:", "$$\n\frac{2y + 3}{y - 2} \leq 0 \quad \ ext{and} \quad \frac{4y - 1}{y - 2} \geq 0\n$$", "Analyze first inequality: $ \frac{2y + 3}{y - 2} \leq 0 $", "Critical points: $ y = -\frac{3}{2} $, $ y = 2 $", "Sign chart:\n- Negative when $ y \in \left[ -\frac{3}{2}, 2 \right) $", "Analyze second inequality: $ \frac{4y - 1}{y - 2} \geq 0 $", "Critical points: $ y = \frac{1}{4} $, $ y = 2 $", "Sign chart:\n- Positive when $ y \in \left( -\infty, \frac{1}{4} \right] \cup (2, \infty) $\n- Undefined at $ y = 2 $", "Find intersection:", "We seek where both inequalities hold:", "- From first: $ y \in \left[ -\frac{3}{2}, 2 \right) $\n- From second: $ y \in \left( -\infty, \frac{1}{4} \right] \cup (2, \infty) $ — but exclude $ y = 2 $", "Intersection is:", "$$\ny \in \left[ -\frac{3}{2}, \frac{1}{4} \right]\n$$", "---", "### Step 5: Final Valid Range for $ y $", "Thus, $ y = \frac{2\sin x + 1}{\sin x - 3} $ must satisfy:", "$$\ny \in \left[ -\frac{3}{2}, \frac{1}{4} \right]\n$$", "This range is essential because it corresponds to real, valid values of $ \sin x $.", "---", "### Step 6: Recover $ \sin x $ in Terms of $ y $", "From earlier:", "$$\n\sin x = \frac{3y + 1}{y - 2}, \quad y \in \left[ -\frac{3}{2}, \frac{1}{4} \right]\n$$", "We can analyze this function over the interval to find its range.", "Let $ f(y) = \frac{3y + 1}{y - 2} $", "Note: $ f(y) $ is continuous and decreasing on $ (-\infty, 2) $, since numerator and denominator are linear and the denominator is nonzero.", "Evaluate $ f(y) $ at endpoints:", "- At $ y = -\frac{3}{2} $:\n $ f\left( -\frac{3}{2} \right) = \frac{3(-\frac{3}{2}) + 1}{-\frac{3}{2} - 2} = \frac{-\frac{9}{2} + 1}{-\frac{7}{2}} = \frac{-\frac{7}{2}}{-\frac{7}{2}} = 1 $", "- At $ y = \frac{1}{4} $:\n $ f\left( \frac{1}{4} \right) = \frac{3 \cdot \frac{1}{4} + 1}{\frac{1}{4} - 2} = \frac{\frac{3}{4} + 1}{\frac{-7}{4}} = \frac{\frac{7}{4}}{-\frac{7}{4}} = -1 $", "Since $ f(y) $ is continuous and strictly decreasing from $ 1 $ to $ -1 $ over $ \left[ -\frac{3}{2}, \frac{1}{4} \right] $, the range of $ \sin x $ is exactly:", "$$\n\sin x \in [-1, 1]\n$$", "But constrained specifically: $ \sin x \in [-1, 1] $ and satisfying the inequality only in the subset $ [-1, \frac{1}{4}] $, which corresponds precisely to:", "$$\n\sin x \in \left[ -1, \frac{1}{4} \right]\n$$", "This confirms that although $ \sin x $ can be in $ [-1,1] $ generally, the expression only validly represents real $ \sin x $ when $ y \in \left[ -\frac{3}{2}, \frac{1}{4} \right] $, matching the domain derived.", "---", "### Conclusion", "By isolating $ \sin x $ from the expression $ y = \frac{2\sin x + 1}{\sin x - 3} $ and imposing the physical constraint $ |\sin x| \leq 1 $, we solve:", "$$\n\sin x = \frac{3y + 1}{y - 2}, \quad y \in \left[ -\frac{3}{2}, \frac{1}{4} \right]\n$$", "This provides a precise, domain-aware solution crucial for equation validity and real-valued trigonometric expressions. Understanding this transformation strengthens problem-solving in advanced trigonometry, calculus, and applied mathematics.", "---", "Keywords:\n$ \sin x = \frac{2\sin x + 1}{\sin x - 3} $, solve for $ \sin x $, trigonometric identities, rational functions, domain constraints, inequality solving, mathematical derivation, algebra with trigonometric functions, science computing, engineering math.", "Meta Description:\nSolve for $ \sin x $ in $ y = \frac{2\sin x + 1}{\sin x - 3} $ by deriving $ \sin x = \frac{3y + 1}{y - 2} $ and find the valid range using $ |\sin x| \leq 1 $. Essential for advanced trigonometry."]