We have

$x^6 - i = 0$

$\therefore x^6 = i \cos \dfrac {\pi}{2}+ i \sin \dfrac {\pi}{2}$

$\therefore x^6 = \cos \left ( 2k\pi + \dfrac {\pi}{2} \right ) + i \sin \left ( 2k \pi + \dfrac {\pi}{2} \right ) $

$\therefore x = \left [ \cos \left ( 2k\pi + \dfrac {\pi}{2} \right ) + i \sin \left ( 2 k \pi + \dfrac {\pi}{2} \right ) \right ]^{\frac {1}{6}}$

Using De Moivre's Theorem,

$\therefore x = \cos (4k +1) \dfrac {\pi}{12} + i \sin (4k + 1) \dfrac {\pi}{12}$

Putting k=0, 1, 2, 3, 4, 5, we get the roots as,

$\cos \left ( \dfrac {\pi}{12} + i \sin \dfrac {\pi}{12} \right ), \cos \left ( \dfrac {5\pi}{12} + i\sin \dfrac {5\pi}{12} \right ), \cos \left ( \dfrac {9\pi}{12} + i \sin \dfrac {9\pi}{12} \right ), \left ( \dfrac {13\pi}{12} + i \sin \dfrac {13\pi}{12} \right ), \cos \left ( \dfrac {17\pi}{12} + i \sin \dfrac {17\pi}{12} \right ), \cos \left ( \dfrac {21\pi}{12} + i \sin \dfrac {21\pi}{12} \right )$

i.e. $\pm \cos \left ( \dfrac {\pi}{12} + i \sin \dfrac {\pi}{12} \right ), \pm \cos \left ( \dfrac {5\pi}{12} + i \sin \dfrac {5\pi}{12} \right ), \pm \cos \left ( \dfrac {9\pi}{12} + i \sin \dfrac {9\pi}{12} \right ), $

i.e. $\pm \cos \left ( \dfrac {3\pi}{12} + i \sin \dfrac {3\pi}{12} \right )$