Step 1 :Define the function \(f(x)=-6 \cos 3x+2\).
Step 2 :Find the derivative of the function, \(f'(x) = 18\sin(3x)\).
Step 3 :Set the derivative equal to zero and solve for \(x\) to find the critical points, \(x = \frac{\pi}{6} + k\pi, \frac{5\pi}{6} + k\pi\) where \(k\) is an integer.
Step 4 :Evaluate the function at the critical points and at the endpoints of the interval \([0, \pi]\) to find the possible maximum and minimum values, \(-4, 8\).
Step 5 :Compare these values to find the maximum and minimum values of the function on the interval.
Step 6 :Final Answer: The minimum value of the function \(f(x)=-6 \cos 3x+2\) on the interval \([0, \pi]\) is \(\boxed{-4}\) and the maximum value is \(\boxed{8}\).