The value of $\alpha$ is in the fourth quadrant, where cosine is negative. Since $\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}$, we have $\cos(\alpha)=-\cos(\frac{2\pi}{3})=-\frac{\sqrt{3}}{2}$. Therefore, $\alpha = -\frac{2\pi}{3}$.