Solution: The equation \(\sin(2x) = \frac{1}{2}\) has solutions when \(2x = \frac{\pi}{6} + 2\pi n\) or \(2x = \frac{5\pi}{6} + 2\pi n\) for integer \(n\). Solving for \(x\) in \([0, 2\pi]\):

["Solving the Equation (\sin(2x) = \frac{1}{2}): A Step-by-Step Guide", "Trigonometric equations often appear challenging at first, but with the right approach, solving them becomes straightforward. One classic example is finding the solutions to the equation:", "[\n\sin(2x) = \frac{1}{2}\n]", "Whether you're a student preparing for exams or tackling math problems independently, understanding how to solve this equation helps build foundational skills in trigonometry.", "---", "### Understanding the Core Equation", "The sine function reaches the value (\frac{1}{2}) at specific angles in the unit circle. We begin by solving the primary angular equations:", "[\n2x = \frac{\pi}{6} + 2\pi n \quad \ ext{or} \quad 2x = \frac{5\pi}{6} + 2\pi n\n]", "where ( n ) is any integer (…,−2, −1, 0, 1, 2, …). These equations capture all possible solutions due to the periodic nature of sine.", "---", "### Step 1: Solve for ( x )", "To isolate ( x ), divide both sides of each equation by 2:", "[\nx = \frac{\pi}{12} + \pi n \quad \ ext{or} \quad x = \frac{5\pi}{12} + \pi n\n]", "Note: Dividing by 2 replaces ( 2x ) with ( x ), and because sine has period ( 2\pi ), adding ( 2\pi n ) inside the sine and then halving gives effective solutions spaced by ( \pi ), not ( 2\pi ).", "---", "### Step 2: Restrict Solutions to the Interval ([0, 2\pi])", "We are interested in all values of ( x ) such that ( 0 \leq x \leq 2\pi ). To find valid ( n ), substitute each solution form into this range.", "#### First family: ( x = \frac{\pi}{12} + \pi n )", "- For ( n = 0 ): ( x = \frac{\pi}{12} ) → within ([0, 2\pi])\n- For ( n = 1 ): ( x = \frac{\pi}{12} + \pi = \frac{13\pi}{12} ) → valid\n- For ( n = 2 ): ( x = \frac{\pi}{12} + 2\pi = \frac{25\pi}{12} ) → greater than ( 2\pi ) → outside range\n- For ( n = -1 ): ( x = \frac{\pi}{12} - \pi = -\frac{11\pi}{12} ) → negative → outside range", "Valid values: ( \frac{\pi}{12}, \frac{13\pi}{12} )", "#### Second family: ( x = \frac{5\pi}{12} + \pi n )", "- For ( n = 0 ): ( x = \frac{5\pi}{12} ) → valid\n- For ( n = 1 ): ( x = \frac{5\pi}{12} + \pi = \frac{17\pi}{12} ) → valid\n- For ( n = 2 ): ( x = \frac{5\pi}{12} + 2\pi = \frac{29\pi}{12} ) >> outside\n- For ( n = -1 ): ( x = \frac{5\pi}{12} - \pi = -\frac{7\pi}{12} ) >> negative", "Valid values: ( \frac{5\pi}{12}, \frac{17\pi}{12} )", "---", "### Final Solutions in ([0, 2\pi])", "Combining both families, the complete solution set is:", "[\nx = \frac{\pi}{12},\ \frac{5\pi}{12},\ \frac{13\pi}{12},\ \frac{17\pi}{12}\n]", "---", "### Why This Method Works", "By solving ( 2x = \frac{\pi}{6} + 2\pi n ) and ( 2x = \frac{5\pi}{6} + 2\pi n ), we account for all angles where sine equals ( \frac{1}{2} ) due to sine's symmetry and periodicity. Dividing by 2 appropriately scales solutions to the variable ( x ), and careful restriction ensures only valid domain values are included.", "---", "### Conclusion", "The equation (\sin(2x) = \frac{1}{2}) has infinitely many solutions, but restricting ( x ) to ([0, 2\pi]) yields a finite set that can be found efficiently using standard trigonometric identities and algebraic manipulation. This method illustrates a powerful approach to solving trigonometric equations systematically—key for mastering precalculus and beyond.", "---", "Key takeaways:", "- Always solve for the inner angle first\n- Account for periodicity and symmetry of sine\n- Restrict solutions to the given domain\n- Use exact fractions and known reference angles", "Mastering these steps will empower you to tackle similar trigonometric equations with confidence. Happy studying!"]