### Instruction: Solve $\cos 3x = 1$ for $0 \le x \le 2 \pi.$  Enter all the solutions, separated by commas.### Response: I know that $\cos \theta = 1$ only when $\theta$ is a multiple of $2 \pi$, so I can write $3x = 2 \pi k$ for some integer $k$.
Then, dividing by 3, I get $x = \frac{2}{3} \pi k$.
Now I need to find all the values of $k$ that make $x$ fall in the interval $[0, 2 \pi]$.
If $k = 0$, then $x = 0$, which is a valid solution.
If $k = 1$, then $x = \frac{2}{3} \pi$, which is also a valid solution.
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, so I'll include it as well.
If $k > 3$, then $x > 2 \pi$, which is outside the interval, so I don't need to consider any larger values of $k$.
Similarly, if $k < 0$, then $x < 0$, which is also outside the interval, so I don't need to consider any smaller values of $k$ either.
Therefore, the only solutions are $x = 0, \frac{2}{3} \pi, \frac{4}{3} \pi, 2 \pi$.