### Instruction: Solve $\cos 3x = 1$ for $0 \le x \le 2 \pi.$  Enter all the solutions, separated by commas.### Response: I know that the cosine function is equal to 1 only when its argument is a multiple of $2 \pi$, so I can write $\cos 3x = 1$ as $3x = 2 \pi k$, where $k$ is any integer.
To solve for $x$, I can divide both sides by 3, and get $x = \frac{2 \pi k}{3}$.
Now I need to find all the values of $k$ that make $x$ satisfy the given interval, $0 \le x \le 2 \pi$.
If I plug in $k = 0$, I get $x = 0$, which is the smallest possible value of $x$.
So $k = 0$ is one solution.
If I increase $k$ by 1, I get $x = \frac{2 \pi}{3}$, which is still in the interval.
So $k = 1$ is another solution.
If I increase $k$ by 1 again, I get $x = \frac{4 \pi}{3}$, which is also in the interval.
So $k = 2$ is another solution.
If I increase $k$ by 1 once more, I get $x = 2 \pi$, which is the largest possible value of $x$.
So $k = 3$ is another solution.
If I increase $k$ by 1 any further, I get $x > 2 \pi$, which is outside the interval.
So $k = 3$ is the largest solution.
Therefore, the solutions are $k = 0, 1, 2, 3$, which correspond to $x = 0, \frac{2 \pi}{3}, \frac{4 \pi}{3}, 2 \pi$.