Precalculus (6th Edition)

$[{cis -30^{\circ},cis -150^{\circ},cis -270^{\circ}}]$
Given: $x^{3}+i=0$ or $x^{3}=-i$ $-i$ can be written in trigonometric form as: $-i=0-i=(\cos 90^{\circ}-\sin 90^{\circ})$ Absolute value of third root is given as $\sqrt[3] 1=1$ Now, the arguments can be given as: $k=0,1,2$ Roots: $(\cos 30^{\circ}+\sin 30^{\circ})$,$(\cos 150^{\circ}+\sin 150^{\circ})$,$(\cos 270^{\circ}+\sin 270^{\circ})$ Solution set of the equation can be written as: $[{cis -30^{\circ},cis -150^{\circ},cis -270^{\circ}}]$