Solution: Solve $ 2\sin(2z) + 1 = 0 \Rightarrow \sin(2z) = -rac12 $. The general solutions for $ 2z $ are $ 2z = 210^\circ + 360^\circ k $ and $ 2z = 330^\circ + 360^\circ k $, where $ k \in \mathbbZ $. 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 $. oxed900 - Abu Waleed Tea