The lab purchased $ \boxed{9} $ units of reagent A.Question: Find all angles $ \theta \in [0^\circ, 360^\circ] $ that satisfy $ \sin(2\theta) = \frac{\sqrt{3}}{2} $.

["Title: Solving $ \sin(2\ heta) = \frac{\sqrt{3}}{2} $: Step-by-Step Guide to All Angles in $ [0^\circ, 360^\circ] $", "When solving trigonometric equations like $ \sin(2\ heta) = \frac{\sqrt{3}}{2} $, understanding how to handle the double-angle form is key. In this article, we’ll find all angles $ \ heta \in [0^\circ, 360^\circ] $ that satisfy the equation $ \sin(2\ heta) = \frac{\sqrt{3}}{2} $, with clear explanations suitable for students, researchers, and lab technicians alike—ideal for deciphering results in chemical experiments requiring precise reagent quantities, such as when purchasing $ 9 $ units of reagent A.", "---", "### Step 1: Let’s Rewrite the Equation", "We start with:", "$$\n\sin(2\ heta) = \frac{\sqrt{3}}{2}\n$$", "We know from the unit circle that $ \sin x = \frac{\sqrt{3}}{2} $ at two standard angles in the interval $ [0^\circ, 360^\circ] $:", "$$\nx = 60^\circ \quad \ ext{and} \quad x = 120^\circ\n$$", "Because sine is positive in Quadrants I and II.", "---", "### Step 2: Substitute $ x = 2\ heta $", "Since the argument of the sine function is $ 2\ heta $, set:", "$$\n2\ heta = 60^\circ \quad \ ext{or} \quad 2\ heta = 120^\circ\n$$", "But sine is periodic with period $ 360^\circ $, so general solutions are:", "$$\n2\ heta = 60^\circ + 360^\circ k \quad \ ext{OR} \quad 2\ heta = 120^\circ + 360^\circ k \quad \ ext{for any integer } k\n$$", "Solving for $ \ heta $:", "$$\n\ heta = 30^\circ + 180^\circ k \quad \ ext{OR} \quad \ heta = 60^\circ + 180^\circ k\n$$", "---", "### Step 3: Find All Solutions in $ [0^\circ, 360^\circ] $", "Now plug in values of $ k $ to find all $ \ heta $ within the desired range.", "#### Case 1: $ \ heta = 30^\circ + 180^\circ k $", "- $ k = 0 $: $ \ heta = 30^\circ $\n- $ k = 1 $: $ \ heta = 210^\circ $\n- $ k = 2 $: $ \ heta = 390^\circ $ → too large", "#### Case 2: $ \ heta = 60^\circ + 180^\circ k $", "- $ k = 0 $: $ \ heta = 60^\circ $\n- $ k = 1 $: $ \ heta = 240^\circ $\n- $ k = 2 $: $ \ heta = 420^\circ $ → too large", "---", "### Step 4: Final Answer – All Valid Angles", "The values of $ \ heta \in [0^\circ, 360^\circ] $ that satisfy $ \sin(2\ heta) = \frac{\sqrt{3}}{2} $ are:", "$$\n\boxed{30^\circ,\ 60^\circ,\ 210^\circ,\ 240^\circ}\n$$", "---", "### Why This Matters in Laboratory Context", "Precision in chemistry and lab work—like when purchasing $ 9 $ units of reagent A—is essential. Understanding angle relationships used in trigonometric models helps verify data precision, especially in spectral analysis or instrument calibration involving rotational or periodic measurements. These solutions may represent critical angles in experimental setups or measurement cycles.", "---", "In summary, solving $ \sin(2\ heta) = \frac{\sqrt{3}}{2} $ yields four key angles between $ 0^\circ $ and $ 360^\circ $: $ 30^\circ, 60^\circ, 210^\circ, 240^\circ $. Using this method ensures accuracy in trigonometric reasoning across science and engineering applications."]