Answer
$\boxed{x = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{9\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}, \frac{21\pi}{12}}$
Work Step by Step
$cos(3x) = sin(3x)$
Let $n =3x$, since $x$ ranges from $0$ to $2\pi$, $n$ ranges from $0$ to $6\pi$.
We have $cos(n) = sin(n)$.
The only values at which $sin$ and $cos$ are equal are multiples of $\frac{\pi}{4}$.
The solutions for $n$ are (which are apparent from looking at a unit circle): $\frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, \frac{13\pi}{4}, \frac{17\pi}{4}, \frac{21\pi}{4}$.
Since we want values of $x$, not $n$, we divide each solution by $3$ because $n=3x$: $x = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{9\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}, \frac{21\pi}{12}$.