Find the maximum and minimum values of the function $g(\theta)=2 \theta-9 \sin (\theta)$ on the interval $[0, \pi]$
Minimum value $=$
Maximum value $=$

Step 1 ：Define the function \(g(\theta)=2 \theta-9 \sin (\theta)\).

Step 2 ：Find the derivative of the function, \(g'(\theta) = 2 - 9\cos(\theta)\).

Step 3 ：Set the derivative equal to zero and solve for \(\theta\) to find the critical points, \(\theta = -\arccos\left(\frac{2}{9}\right) + 2\pi, \arccos\left(\frac{2}{9}\right)\).

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, \(-2\arccos\left(\frac{2}{9}\right) + \sqrt{77} + 4\pi, -\sqrt{77} + 2\arccos\left(\frac{2}{9}\right), 0, 2\pi\).

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 \(g(\theta)=2 \theta-9 \sin (\theta)\) on the interval \([0, \pi]\) is \(\boxed{-\sqrt{77} + 2\arccos\left(\frac{2}{9}\right)}\) and the maximum value is \(\boxed{-2\arccos\left(\frac{2}{9}\right) + \sqrt{77} + 4\pi}\).