To solve the equation \(\cot(3x + 30^\circ) = -\frac{\sqrt{3}}{3}\), we need to find the angle whose cotangent is \(-\frac{\sqrt{3}}{3}\).

First, recall that \(\cot(\theta) = \frac{1}{\tan(\theta)}\). Therefore, we need to find the angle \(\theta\) such that \(\tan(\theta) = -\sqrt{3}\).

We know that \(\tan(30^\circ) = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}\). Therefore, \(\tan(150^\circ) = -\frac{\sqrt{3}}{3}\) because tangent is negative in the second quadrant.

So, we have:
\[ 3x + 30^\circ = 150^\circ + k \cdot 180^\circ \]
where \(k\) is any integer, because the tangent function has a period of \(180^\circ\).

Now, solve for \(x\):
\[ 3x + 30^\circ = 150^\circ + k \cdot 180^\circ \]
\[ 3x = 150^\circ - 30^\circ + k \cdot 180^\circ \]
\[ 3x = 120^\circ + k \cdot 180^\circ \]
\[ x = 40^\circ + k \cdot 60^\circ \]

Therefore, the general solution for \(x\) is:
\[ x = 40^\circ + k \cdot 60^\circ \]
where \(k\) is any integer.