#### Answer

$[{cis -30^{\circ},cis -150^{\circ},cis -270^{\circ}}]$

#### Work Step by Step

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:
$[{cis -30^{\circ},cis -150^{\circ},cis -270^{\circ}}]$