Solution: Solve $2z = 60^\circ, 120^\circ, 420^\circ, 480^\circ$ within $[0^\circ, 720^\circ]$, yielding $

["Solving $2z = 60^\circ, 120^\circ, 420^\circ, 480^\circ$ in the Interval $[0^\circ, 720^\circ]$: A Step-by-Step Guide", "When solving trigonometric equations involving an angle multiplied by a constant, it’s essential to carefully determine all valid solutions within the specified domain. In this article, we’ll solve the equation $2z = 60^\circ, 120^\circ, 420^\circ, 480^\circ$ and find all values of $z$ that lie within the interval $[0^\circ, 720^\circ]$.", "---", "### Understanding the Equation", "We are given:\n$$\n2z = 60^\circ, 120^\circ, 420^\circ, 480^\circ\n$$\nTo isolate $z$, divide each right-hand side by 2:\n$$\nz = \frac{60^\circ}{2},\quad z = \frac{120^\circ}{2},\quad z = \frac{420^\circ}{2},\quad z = \frac{480^\circ}{2}\n$$", "---", "### Step-by-Step Calculation", "Let’s compute each value:", "1.\n$$\nz = \frac{60^\circ}{2} = 30^\circ\n$$", "2.\n$$\nz = \frac{120^\circ}{2} = 60^\circ\n$$", "3.\n$$\nz = \frac{420^\circ}{2} = 210^\circ\n$$", "4.\n$$\nz = \frac{480^\circ}{2} = 240^\circ\n$$", "So far, we have the candidate solutions:\n$$\nz = 30^\circ,\ 60^\circ,\ 210^\circ,\ 240^\circ\n$$", "---", "### Checking the Domain $[0^\circ, 720^\circ]$", "All the computed values—$30^\circ, 60^\circ, 210^\circ, 240^\circ$—are less than $720^\circ$, but we must ensure no value exceeds $720^\circ$, especially when considering periodicity or repeated angles. However, since we directly solved for $z$ and the original angles $60^\circ, 120^\circ, 420^\circ, 480^\circ$ are all in $[0^\circ, 720^\circ]$, and $2z$ yields exactly those, and $z = \frac{\ ext{angle}}{2}$ preserves the domain, we now confirm no additional solutions emerge from periodic repetition within the interval.", "Note: The equation $2z = \ heta$ gives a unique solution for $z$ per angle $\ heta$. Since all four $\ heta$ values are in the basic range and $z$ values remain under $720^\circ$, no further solutions appear.", "---", "### Final Answer", "The solutions to $2z = 60^\circ, 120^\circ, 420^\circ, 480^\circ$ in the interval $[0^\circ, 720^\circ]$ are:\n$$\n\boxed{30^\circ,\ 60^\circ,\ 210^\circ,\ 240^\circ}\n$$", "---", "### Bonus Tip: General Strategy", "For equations of the form $kz = \ heta$ with $k <br/>\ne 0$, solve for $z = \frac{\ heta}{k}$. Then find all angles $\ heta + 360^\circ n$ divided by $k$, for integer $n$, such that $z \in [0^\circ, K]$, where $K = 720^\circ$. This ensures all periodic solutions within bounds are captured.", "---", "Keywords: solve $2z = 60^\circ$, solve $2z = 120^\circ$, solve $2z = 420^\circ$, solve $2z = 480^\circ$, find $z$ in $[0^\circ, 720^\circ]$, trigonometric equation solution, step-by-step algebra"]