Question: Determine the number of real solutions to the equation \(\sin(2x) = \frac{1}{2}$ for \(x \in [0, 2\pi]\).

["Title: Determine the Number of Real Solutions to (\sin(2x) = \frac{1}{2}) in the Interval ([0, 2\pi])", "---", "### Introduction", "Understanding how to determine the number of real solutions to trigonometric equations is essential in calculus, algebra, and advanced math studies. One particularly common problem involves solving equations of the form (\sin(2x) = \frac{1}{2}) over a specific interval, such as ([0, 2\pi]). This article walks through the steps to determine how many real solutions exist for this equation, combining graph analysis, periodic properties, and algebraic techniques.", "---", "### Understanding the Equation: (\sin(2x) = \frac{1}{2})", "The equation (\sin(2x) = \frac{1}{2}) asks for all angles (2x) such that the sine value equals ( \frac{1}{2} ). Since the argument (2x) appears in the function, it determines a scaled horizontal input compared to the standard sine function.", "We begin by solving the simpler trigonometric equation:", "[\n\sin(\ heta) = \frac{1}{2}\n]", "The general real solutions for (\ heta) are:", "[\n\ heta = \frac{\pi}{6} + 2k\pi \quad \ ext{and} \quad \ heta = \frac{5\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}\n]", "---", "### Substituting Back: Solve for (x)", "Since (\ heta = 2x), we substitute into the general solution:", "[\n2x = \frac{\pi}{6} + 2k\pi \quad \Rightarrow \quad x = \frac{\pi}{12} + k\pi\n]", "and", "[\n2x = \frac{5\pi}{6} + 2k\pi \quad \Rightarrow \quad x = \frac{5\pi}{12} + k\pi\n]", "Thus, the two families of solutions are:", "- (x = \frac{\pi}{12} + k\pi)\n- (x = \frac{5\pi}{12} + k\pi), for all integers (k)", "---", "### Restricting the Domain: (x \in [0, 2\pi])", "We now determine the values of (k) such that each expression stays within ([0, 2\pi]).", "#### First family: (x = \frac{\pi}{12} + k\pi)", "- For (k = 0): (x = \frac{\pi}{12} \approx 0.2618) → valid\n- For (k = 1): (x = \frac{\pi}{12} + \pi = \frac{13\pi}{12} \approx 3.4034) → valid\n- For (k = 2): (x = \frac{\pi}{12} + 2\pi = \frac{25\pi}{12} \approx 6.5452 > 2\pi) → invalid", "So only (k = 0, 1) yield valid solutions in ([0, 2\pi]):", "[\nx = \frac{\pi}{12}, \quad x = \frac{13\pi}{12}\n]", "#### Second family: (x = \frac{5\pi}{12} + k\pi)", "- For (k = 0): (x = \frac{5\pi}{12} \approx 1.308) → valid\n- For (k = 1): (x = \frac{5\pi}{12} + \pi = \frac{17\pi}{12} \approx 4.4506) → valid\n- For (k = 2): (x = \frac{5\pi}{12} + 2\pi = \frac{29\pi}{12} > 2\pi) → invalid", "So valid solutions are:", "[\nx = \frac{5\pi}{12}, \quad x = \frac{17\pi}{12}\n]", "---", "### Summary of Solutions", "Combining both families within ([0, 2\pi]), the real solutions are:", "[\nx = \frac{\pi}{12},\ \frac{5\pi}{12},\ \frac{13\pi}{12},\ \frac{17\pi}{12}\n]", "Thus, there are exactly 4 real solutions.", "---", "### Why This Method Works: Periodicity and Domain Restriction", "The sine function is periodic with period (2\pi), but here the argument (2x) compresses the period: the function (\sin(2x)) completes one full cycle over an interval of length (\pi). This means the sine wave completes two full oscillations over ([0, 2\pi]), resulting in more solutions.", "Because we restricted (x) to a finite interval, only 4 distinct solutions satisfy both the equation and the domain.", "---", "### Graphical Insight", "Plotting (y = \sin(2x)) over ([0, 2\pi]) reveals two complete sine waves. The horizontal lines (y = \frac{1}{2}) intersect the curve at four points—two per cycle—confirming our algebraic result.", "---", "### Conclusion", "To determine the number of real solutions to (\sin(2x) = \frac{1}{2}) for (x \in [0, 2\pi]):", "1. Solve the basic sine equation to find all general solutions.\n2. Substitute the transformation (2x) to adapt the angles.\n3. Restrict each solution family to the domain ([0, 2\pi]) by testing integer values of (k).\n4. Count distinct valid (x)-values.", "Using this method, we find there are exactly four real solutions.", "---", "### Key Takeaway", "Mastering trigonometric equation solving involves recognizing periodic behavior, applying identity substitutions, and carefully restricting domains. Consistent practice with real-world contexts like oscillatory motion strengthens fluency in analyzing such equations.", "---", "Keywords: (\sin(2x) = \frac{1}{2}), number of solutions, real solutions, trigonometric equation, interval ([0, 2\pi]), sine period, graphical analysis, algebra, trigonometric identities.", "Meta Description:\nDiscover how to determine the number of real solutions to (\sin(2x) = \frac{1}{2}) for (x \in [0, 2\pi]) using general solutions, domain restrictions, and periodicity. Learn step-by-step with examples and practice insights.", "---", "Happy solving! For further depth, explore similar equations with different amplitudes or modulated arguments."]