الحل: لحل المعادلة $\sin(2z) = \frac{\sqrt{3}}{2}$، نحدد أولاً الزوايا التي تساوي فيها دالة الجيب $\frac{\sqrt{3}}{2}$. هذه الزوايا في الربعين الأول والثاني ضمن الفترة $[0^\circ, 360^\circ]$ هي $60^\circ$ و $120^\circ$. بما أن وسيط دالة الجيب ضمن المعادلة هو $2z$، فإننا نحل لإيجاد $2z$:

["The Solution: Solving the Equation $\sin(2z) = \frac{\sqrt{3}}{2}$", "When solving trigonometric equations like $\sin(2z) = \frac{\sqrt{3}}{2}$, the key is to first identify all angles within the fundamental period where the sine function takes this value, then adjust for the specific variable $2z$, and finally solve for $z$.", "---", "### Step 1: Identify Reference Angles", "The sine function equals $\frac{\sqrt{3}}{2}$ at specific angles. Within $[0^\circ, 360^\circ]$, these occur at:", "$$\n\sin(\ heta) = \frac{\sqrt{3}}{2} \implies \ heta = 60^\circ \quad \ ext{and} \quad \ heta = 120^\circ\n$$", "These are the reference angles in the first and second quadrants where sine is positive.", "---", "### Step 2: Solve for $2z$", "In our equation $\sin(2z) = \frac{\sqrt{3}}{2}$, the variable inside the sine is $2z$. So we solve:", "$$\n2z = 60^\circ \quad \ ext{or} \quad 2z = 120^\circ\n$$", "But since sine is periodic with a period of $360^\circ$, we must include all coterminal solutions within a suitable interval. However, since $z$ is typically considered in degrees across $0^\circ \leq z < 360^\circ$, then $2z$ ranges from $0^\circ$ to $720^\circ$. We find all solutions of $\sin(\ heta) = \frac{\sqrt{3}}{2}$ for $\ heta = 2z$ in $[0^\circ, 720^\circ]$.", "The general solutions for $\sin(\ heta) = \frac{\sqrt{3}}{2}$ are:", "$$\n\ heta = 60^\circ + 360^\circ k \quad \ ext{and} \quad \ heta = 120^\circ + 360^\circ k \quad \ ext{for integer } k\n$$", "We now find all such $\ heta$ in $[0^\circ, 720^\circ]$:", "- For $k = 0$:\n $\ heta = 60^\circ$, $120^\circ$", "- For $k = 1$:\n $\ heta = 420^\circ$ ($60^\circ + 360^\circ$), $480^\circ$ ($120^\circ + 360^\circ$)", "- For $k = 2$:\n $\ heta = 780^\circ > 720^\circ$ → too large", "So the valid values of $2z$ are:", "$$\n2z = 60^\circ, ; 120^\circ, ; 420^\circ, ; 480^\circ\n$$", "---", "### Step 3: Solve for $z$", "Now divide each solution by 2 to find $z$:", "$$\nz = \frac{60^\circ}{2} = 30^\circ\n$$\n$$\nz = \frac{120^\circ}{2} = 60^\circ\n$$\n$$\nz = \frac{420^\circ}{2} = 210^\circ\n$$\n$$\nz = \frac{480^\circ}{2} = 240^\circ\n$$", "All four values lie within $[0^\circ, 360^\circ)$, so they are valid.", "---", "### Final Answer:", "$$\n\boxed{z = 30^\circ,\ 60^\circ,\ 210^\circ,\ 240^\circ}\n$$", "---", "### Summary", "To solve $\sin(2z) = \frac{\sqrt{3}}{2}$:\n1. Identify the reference angles: $60^\circ$ and $120^\circ$.\n2. Solve $2z = 60^\circ + 360^\circ k$ and $2z = 120^\circ + 360^\circ k$.\n3. For $k = 0, 1$, found $2z = 60^\circ, 120^\circ, 420^\circ, 480^\circ$.\n4. Divide each by 2: $z = 30^\circ, 60^\circ, 210^\circ, 240^\circ$.", "This method ensures all solutions are captured efficiently using periodicity and domain considerations.", "# Trigonometry # Equation Solving # $\sin(2z) = \frac{\sqrt{3}}{2}$ # Step-by-Step Solutions # Math Tips"]