Solution: Let $ y = \sin(2x) $. The equation becomes $ y^2 + 3y + 2 = 0 $, which factors as $ (y + 1)(y + 2) = 0 $. Thus, $ y = -1 $ or $ y = -2 $. Since $ \sin(2x) $ ranges between $-1$ and $1$, $ y = -2 $ is invalid. For $ y = -1 $, $ \sin(2x) = -1 $ has infinitely many solutions (e.g., $ 2x = rac{3\pi}{2} + 2\pi k $, $ k \in \mathbb{Z} $). However, if restricted to a specific interval (not stated here), the count would depend on the domain. Assuming $ x \in \mathbb{R} $, there are infinitely
["Solving $ \sin(2x) = y $: Understanding Solutions When $ y^2 + 3y + 2 = 0 $", "Mathematics often involves transforming complex equations into simpler forms to reveal elegant solutions. One insightful approach is recognizing when a substitution reduces a trigonometric equation to an algebraic one. Here, we explore the equation $ \sin(2x) = y $, which transforms into the quadratic $ y^2 + 3y + 2 = 0 $. This simple yet powerful substitution unlocks a clear path to finding valid solutions for $ x $.", "### Step 1: Substitute and Simplify", "Let $ y = \sin(2x) $. Substituting into the original equation gives:", "$$\ny^2 + 3y + 2 = 0\n$$", "This is a standard quadratic equation. Factoring yields:", "$$\n(y + 1)(y + 2) = 0\n$$", "Thus, the potential solutions are:", "$$\ny = -1 \quad \ ext{or} \quad y = -2\n$$", "### Step 2: Analyze Validity Using Trigonometric Boundaries", "Recall that $ \sin(2x) $ is bounded between $-1$ and $1$:", "$$\n-1 \leq \sin(2x) \leq 1\n$$", "This means that $ y = -2 $ lies outside this range and cannot correspond to any valid sine value. Only:", "$$\ny = -1\n$$", "is a feasible solution.", "### Step 3: Solve for $ x $ When $ \sin(2x) = -1 $", "We now solve:", "$$\n\sin(2x) = -1\n$$", "The general solution for $ \sin \ heta = -1 $ is:", "$$\n\ heta = \frac{3\pi}{2} + 2\pi k, \quad k \in \mathbb{Z}\n$$", "Substitute $ \ heta = 2x $:", "$$\n2x = \frac{3\pi}{2} + 2\pi k\n$$", "Solving for $ x $:", "$$\nx = \frac{3\pi}{4} + \pi k\n$$", "### Step 4: Determine the Number of Solutions", "Given $ x \in \mathbb{R} $, the expression $ x = \frac{3\pi}{4} + \pi k $ produces infinitely many distinct solutions—each differing by $ \pi $. There is no upper or lower bound on $ k $, so the number of solutions is infinite.", "> 📝 Note: If the problem had specified a particular interval (e.g., $ x \in [0, 2\pi) $), the number of solutions would be finite. However, without such restriction, the solutions extend infinitely across the real line.", "### Final Answer", "The equation $ \sin(2x) = y $ transforms via $ y^2 + 3y + 2 = 0 $ to $ y = -1 $ as the only valid solution within the range of sine, yielding infinitely many real solutions:", "$$\nx = \frac{3\pi}{4} + \pi k, \quad k \in \mathbb{Z}\n$$", "Therefore, the total number of solutions is:", "$$\n\boxed{\ ext{infinitely many}}\n$$", "This example beautifully demonstrates how algebraic manipulation can simplify trigonometric equations—and how domain constraints are crucial for identifying valid solutions."]
