Solve the given equation. (Enter your answers as a comma-separated list. Let \( k \) be any integer.) \[ \tan (\theta)=-\frac{\sqrt{3}}{3} \] \[ \theta= \] \( \square \) rad

To solve the equation \( \tan(\theta) = -\frac{\sqrt{3}}{3} \), we start by identifying the reference angle where the tangent value is \(\frac{\sqrt{3}}{3}\). This occurs at \(\theta = \frac{\pi}{6}\) radians. However, since we have a negative tangent, we need to look for angles in the second and fourth quadrants, where tangent is negative. In the second quadrant, we find: \[ \theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}. \] In the fourth quadrant: \[ \theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}. \] Now, since the tangent function has a period of \(\pi\), we can express the general solutions as: \[ \theta = \frac{5\pi}{6} + k\pi, \quad \frac{11\pi}{6} + k\pi, \] where \( k \) is any integer. So, the answers are: \[ \theta = \frac{5\pi}{6} + k\pi, \frac{11\pi}{6} + k\pi. \]