Solution: By De Moivre's Theorem, $z^6 = \cos(6\theta) + i\sin(6\theta) = -1 + 0i$. This implies $6\theta = \pi + 2\pi k$ for integer $k$. Solving for $\theta$ gives $\theta = \frac\pi6 + \frac\pi k3$. The principal solution in $[0, 2\pi)$ is $\theta = \frac\pi6, \frac\pi2, \frac5\pi6, \frac7\pi6, \frac3\pi2, \frac11\pi6$. The smallest positive $\theta$ is $\boxed\dfrac\pi6$. - Portal da Acústica