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 \( 0 \leq x < 2\pi \).
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 \( 0 \leq x < 2\pi \). Calculate these values to find all possible solutions for \( x \).