### Instruction: Solve $\cos 3x = 1$ for $0 \le x \le 2 \pi.$  Enter all the solutions, separated by commas.### Response: To solve this equation, I need to find the angles that have a cosine value of 1.
I know that cosine is 1 when the angle is 0 or a multiple of $2 \pi.$
So I can write $\cos 3x = 1$ as $3x = 2k \pi$ for some integer $k.$
To isolate $x,$ I can divide both sides by 3, and get $x = \frac{2}{3} k \pi.$
Now I need to find the values of $k$ that make $x$ fall within the given interval of $0 \le x \le 2 \pi.$
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 within the interval.
If $k = 3,$ then $x = 2 \pi,$ which is the upper bound of the interval and also a valid solution.
If $k > 3,$ then $x > 2 \pi,$ which is outside the interval and not a valid solution.
If $k < 0,$ then $x < 0,$ which is also outside the interval and not a valid solution.
Therefore, the only possible values of $k$ are 0, 1, 2, and 3, and the corresponding values of $x$ are the solutions of the equation.