Solution: Solve $ 2\sin(2z) + 1 = 0 \Rightarrow \sin(2z) = -rac{1}{2} $. The general solutions for $ 2z $ are $ 2z = 210^\circ + 360^\circ k $ and $ 2z = 330^\circ + 360^\circ k $, where $ k \in \mathbb{Z} $. Dividing by 2, $ z = 105^\circ, 165^\circ, 285^\circ, 345^\circ $ within $ [0^\circ, 360^\circ] $. The sum is $ 105 + 165 + 285 + 345 = 900^\circ $. oxed{900}

["Solve $ 2\sin(2z) + 1 = 0 $: General Solutions and the Sum of All Solutions in $[0^\circ, 360^\circ]$", "Trigonometric equations often challenge students and math enthusiasts alike. Today, we explore a clean and insightful solution to the equation:\n[\n2\sin(2z) + 1 = 0 \quad \Rightarrow \quad \sin(2z) = -\frac{1}{2}\n]", "---", "Step 1: Simplify the Equation", "We begin by isolating the sine term:\n[\n\sin(2z) = -\frac{1}{2}\n]", "Let $ \ heta = 2z $. The equation becomes:\n[\n\sin(\ heta) = -\frac{1}{2}\n]", "---", "Step 2: Find General Solutions for $ \ heta $", "Recall that $ \sin(\ heta) = -\frac{1}{2} $ occurs at standard angles in the third and fourth quadrants. The reference angle for $ \sin^{-1}\left(\frac{1}{2}\right) $ is $ 30^\circ $. Therefore, the general solutions for $ \ heta $ are:\n[\n\ heta = 210^\circ + 360^\circ k \quad \ ext{and} \quad \ heta = 330^\circ + 360^\circ k \quad \ ext{where } k \in \mathbb{Z}\n]", "---", "Step 3: Back-substitute $ \ heta = 2z $", "Now replace $ \ heta $ with $ 2z $:\n[\n2z = 210^\circ + 360^\circ k \quad \Rightarrow \quad z = 105^\circ + 180^\circ k\n]\n[\n2z = 330^\circ + 360^\circ k \quad \Rightarrow \quad z = 165^\circ + 180^\circ k\n]", "---", "Step 4: Find All $ z $ in $[0^\circ, 360^\circ]$", "We substitute integer values of $ k $ to find all solutions within the interval:", "For $ z = 105^\circ + 180^\circ k $:\n- $ k = 0 \Rightarrow z = 105^\circ $\n- $ k = 1 \Rightarrow z = 285^\circ $\n- $ k = -1 $ gives $ z < 0^\circ $, out of range", "For $ z = 165^\circ + 180^\circ k $:\n- $ k = 0 \Rightarrow z = 165^\circ $\n- $ k = 1 \Rightarrow z = 345^\circ $\n- $ k = -1 $ gives $ z < 0^\circ $, out of range", "So the complete set of solutions in $[0^\circ, 360^\circ]$ is:\n[\nz = 105^\circ, 165^\circ, 285^\circ, 345^\circ\n]", "---", "Step 5: Calculate the Sum of All Solutions", "Add the solutions:\n[\n105^\circ + 165^\circ + 285^\circ + 345^\circ = 900^\circ\n]", "---", "Why This Method Works", "By solving for the angle’s general behavior first and then carefully restricting the variable, we avoid missing solutions while keeping the process transparent and systematic. Understanding the periodicity of sine ($360^\circ$) and the doubling variable ($2z$) is key.", "---", "Final Answer:\n[\n\boxed{900}\n]"]