### Instruction: Solve $\cos 3x = 1$ for $0 \le x \le 2 \pi.$  Enter all the solutions, separated by commas.### Response: I know that $\cos \theta = 1$ when $\theta$ is a multiple of $2 \pi,$ so I want to find all the values of $3x$ that are multiples of $2 \pi$ in the given interval.
I can write $3x = 2 \pi k,$ where $k$ is an integer, and then divide both sides by $3$ to get $x = \frac{2}{3} \pi k.$
Now I need to find the values of $k$ that make $x$ between $0$ and $2 \pi.$
If $k = 0,$ then $x = 0,$ which is in the interval.
If $k = 1,$ then $x = \frac{2}{3} \pi,$ which is also in the interval.
If $k = 2,$ then $x = \frac{4}{3} \pi,$ which is still in the interval.
If $k = 3,$ then $x = 2 \pi,$ which is the endpoint of the interval.
If $k > 3,$ then $x > 2 \pi,$ which is outside the interval.
If $k < 0,$ then $x < 0,$ which is also outside the interval.
Therefore, the only solutions are $x = 0, \frac{2}{3} \pi, \frac{4}{3} \pi, 2 \pi.$