Answer
$\frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2} $
Work Step by Step
Step 1. Given the equation $sin(3x)=1$, we can find the solution in $[0,2\pi)$ as $3x=\frac{\pi}{2}$ or $x=\frac{\pi}{6}$
Step 2. Consider the original function has a period of $\frac{2\pi}{3}$; we can express all the solutions as $x=\frac{2\pi}{3}k+\frac{\pi}{6}$ where $k$ is an integer. Thus, within $[0,2\pi)$, we have $x=\frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}$ as solutions to the original equation.
