Solve the following equations.
$\sin3x=\frac{\sqrt{2}}{2},0\leq{x}\lt{2\pi}$

Verified step by step guidance

1. First, recognize that the equation $\sin (3x)=\frac{\sqrt{2}}{2}$ is a trigonometric equation where we need to find the values of $x$ within the interval $0\le x<2\pi$.
2. Recall that $\sin (\theta )=\frac{\sqrt{2}}{2}$ corresponds to angles $\theta =\frac{\pi }{4}$ and $\theta =\frac{3\pi }{4}$ in the unit circle.
3. Since we have $\sin (3x)=\frac{\sqrt{2}}{2}$, set $3x=\frac{\pi }{4}+2k\pi$ and $3x=\frac{3\pi }{4}+2k\pi$ for integer values of $k$.
4. Solve for $x$ by dividing each equation by 3: $x=\frac{\pi }{12}+\frac{2k\pi }{3}$ and $x=\frac{\pi }{4}+\frac{2k\pi }{3}$.
5. Determine the values of $k$ such that $x$ remains within the interval $0\le x<2\pi$. Calculate these values to find all possible solutions for $x$.