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 $\theta = \frac{\pi}{6} + \frac{\pi k}{3}$. The principal solution in $[0, 2\pi)$ is $\theta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}$. The smallest positive $\theta$ is $\boxed{\dfrac{\pi}{6}}$.

["Solving $z^6 = -1$ Using De Moivre’s Theorem: Find All Solutions in $[0, 2\pi)$", "When solving complex equations involving powers of complex numbers, De Moivre’s Theorem provides a powerful approach. One classic example is finding all sixth roots of $-1$, where the expression $z^6 = -1$ reveals elegant geometric and algebraic insights.", "---", "### Using De Moivre’s Theorem to Solve $z^6 = -1$", "De Moivre’s Theorem states that $(\cos\ heta + i\sin\ heta)^n = \cos(n\ heta) + i\sin(n\ heta)$. Since $-1$ can be written in polar form as $\cos\pi + i\sin\pi$, we express the equation as:", "$$\nz^6 = \cos\pi + i\sin\pi\n$$", "By De Moivre’s formula, the general solution for $z$ is:", "$$\nz = \cos\left(\frac{\pi + 2\pi k}{6}\right) + i\sin\left(\frac{\pi + 2\pi k}{6}\right), \quad \ ext{for } k = 0, 1, 2, 3, 4, 5\n$$", "Simplifying the angle:", "$$\n\ heta_k = \frac{\pi}{6} + \frac{\pi k}{3}\n$$", "---", "### Finding All Solutions in $[0, 2\pi)$", "Substitute integer values of $k$ from 0 to 5 to find all distinct solutions in the principal range:", "- For $k = 0$: $\ heta = \frac{\pi}{6}$\n- For $k = 1$: $\ heta = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$\n- For $k = 2$: $\ heta = \frac{\pi}{6} + \frac{2\pi}{3} = \frac{5\pi}{6}$\n- For $k = 3$: $\ heta = \frac{\pi}{6} + \pi = \frac{7\pi}{6}$\n- For $k = 4$: $\ heta = \frac{\pi}{6} + \frac{4\pi}{3} = \frac{3\pi}{2}$\n- For $k = 5$: $\ heta = \frac{\pi}{6} + \frac{5\pi}{3} = \frac{11\pi}{6}$", "All six angles lie within $[0, 2\pi)$, giving the full set of roots:", "$$\n\ heta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}\n$$", "---", "### Identify the Smallest Positive Solution", "Among these, the smallest positive value of $\ heta$ is clearly the principal one:", "$$\n\boxed{\dfrac{\pi}{6}}\n$$", "This angle represents the first sixth root of $-1$ in the complex plane, demonstrating how De Moivre’s Theorem transforms algebraic problems into intuitive geometric solutions.", "---", "### Conclusion", "De Moivre’s Theorem simplifies solving $z^n = r(\cos\ heta + i\sin\ heta)$ by converting powers into angular multiplications. Applying this to $z^6 = -1$, we identify all six distinct roots in $[0, 2\pi)$, with the smallest positive solution being $\ heta = \frac{\pi}{6}$. This method is invaluable in complex analysis, signal processing, and engineering applications involving rotational symmetry and periodic behavior."]