Now solve $ \sin(2x) = \frac{\sqrt{2}}{2} $ for $ x \in (0, \pi) $. Then $ 2x \in (0, 2\pi) $.

["Solving the Equation $ \sin(2x) = \frac{\sqrt{2}}{2} $ for $ x \in (0, \pi) $ — A Step-by-Step Guide", "Trigonometric equations like $ \sin(2x) = \frac{\sqrt{2}}{2} $ frequently appear in mathematics, physics, and engineering. Understanding how to solve such equations is essential for anyone studying advanced algebra, calculus, or applied sciences. In this article, we’ll solve $ \sin(2x) = \frac{\sqrt{2}}{2} $ with the domain $ x \in (0, \pi) $, noting that this implies $ 2x \in (0, 2\pi) $.", "---", "### Understanding the Equation", "The equation $ \sin(2x) = \frac{\sqrt{2}}{2} $ asks: For what values of $ 2x $ in $ (0, 2\pi) $ does the sine function equal $ \frac{\sqrt{2}}{2} $? Recall that:", "$$\n\frac{\sqrt{2}}{2} = \sin\left(\frac{\pi}{4}\right) = \sin\left(\frac{3\pi}{4}\right)\n$$", "These are two distinct angles in $ (0, 2\pi) $ where sine equals $ \frac{\sqrt{2}}{2} $. Because sine is positive in both the first and second quadrants, we use:", "- $ 2x = \frac{\pi}{4} $\n- $ 2x = \frac{3\pi}{4} $", "These are the principal solutions to $ \sin \ heta = \frac{\sqrt{2}}{2} $ in $ (0, 2\pi) $.", "---", "### Solving for $ x $", "Since $ 2x \in (0, 2\pi) $, we solve for $ x $:", "1. $ 2x = \frac{\pi}{4} \Rightarrow x = \frac{\pi}{8} $\n2. $ 2x = \frac{3\pi}{4} \Rightarrow x = \frac{3\pi}{8} $", "Are there any other solutions?\nThe sine function has a period of $ 2\pi $, but $ 2x \in (0, 2\pi) $, so $ x \in (0, \pi) $ covers all relevant values. Between $ 0 $ and $ 2\pi $, $ \sin \ heta = \frac{\sqrt{2}}{2} $ has only two solutions — $ \frac{\pi}{4} $ and $ \frac{3\pi}{4} $. Therefore, only two solutions exist in $ (0, \pi) $.", "---", "### Verifying the Solutions", "Let’s verify both values:", "- For $ x = \frac{\pi}{8} $: $ 2x = \frac{\pi}{4} \Rightarrow \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} $ ✓\n- For $ x = \frac{3\pi}{8} $: $ 2x = \frac{3\pi}{4} \Rightarrow \sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2} $ ✓", "Both satisfy the equation.", "---", "### Final Answer", "The solutions to $ \sin(2x) = \frac{\sqrt{2}}{2} $ in the interval $ x \in (0, \pi) $ are:", "$$\n\boxed{x = \frac{\pi}{8} \quad \ ext{and} \quad x = \frac{3\pi}{8}}\n$$", "---", "### Why This Matters", "Solving trigonometric equations in bounded domains like $ x \in (0, \pi) $ appears in wave analysis, harmonic motion, and signal processing. Mastering this foundational skill supports deeper exploration of periodic functions and their real-world applications.", "If you're studying similar equations, remember:\n- Identify the inner angle.\n- Find all solutions in the correct interval.\n- Apply periodicity only when extending beyond one cycle.", "Happy solving! 📐"]