Answer
$$x = \frac{1}{{12}}\pi ,{\text{ }}\frac{1}{4}\pi ,{\text{ }}\frac{3}{4}\pi ,{\text{ }}\frac{{17}}{{12}}\pi ,{\text{ }}\frac{{19}}{{12}}\pi $$
Work Step by Step
$$\eqalign{
& \sin 3x = \frac{{\sqrt 2 }}{2},\,\,\,\,\,0 \leqslant x < 2\pi \cr
& {\text{In the interval }}\left[ {0,2\pi } \right),{\text{ sin}}\theta = \frac{{\sqrt 2 }}{2}{\text{ at }}\theta = \frac{\pi }{4}{\text{ and }}\theta = \frac{{3\pi }}{4} \cr
& {\text{Therefore}}{\text{, Let }}\theta = 3x \cr
& 3x = \frac{\pi }{4}{\text{ and }}3x = \frac{{3\pi }}{4} \cr
& x = \frac{\pi }{{12}}{\text{ and }}x = \frac{\pi }{4} \cr
& {\text{The general solution is}} \cr
& x = \frac{\pi }{{12}} + \frac{{2n\pi }}{3}{\text{ and }}x = \frac{\pi }{4} + \frac{{2n\pi }}{3},{\text{ }}n{\text{ is an integer}} \cr
& x = \left( {\frac{{1 + 8n}}{{12}}} \right)\pi {\text{ and }}x = \left( {\frac{{3 + 8n}}{{12}}} \right)\pi \cr
& {\text{For }}n = 0 \cr
& x = \frac{1}{{12}}\pi {\text{ and }}x = \frac{1}{4}\pi ,{\text{ }}\left( {{\text{are on the interval }}\,0 \leqslant x < 2\pi } \right) \cr
& {\text{For }}n = 1 \cr
& x = \frac{3}{4}\pi {\text{ and }}x = \frac{{11}}{{12}}\pi ,{\text{ }}\left( {{\text{are on the interval }}\,0 \leqslant x < 2\pi } \right) \cr
& {\text{For }}n = 2 \cr
& x = \frac{{17}}{{12}}\pi {\text{ and }}x = \frac{{19}}{{12}}\pi ,{\text{ }}\left( {{\text{are on the interval }}\,0 \leqslant x < 2\pi } \right) \cr
& {\text{For }}n = 3 \cr
& x = \frac{{25}}{{12}}\pi {\text{ and }}x = \frac{9}{4}\pi ,{\text{ }}\left( {{\text{are not on the interval }}\,0 \leqslant x < 2\pi } \right) \cr
& {\text{The solutions are:}} \cr
& x = \frac{1}{{12}}\pi ,{\text{ }}\frac{1}{4}\pi ,{\text{ }}\frac{3}{4}\pi ,{\text{ }}\frac{{17}}{{12}}\pi ,{\text{ }}\frac{{19}}{{12}}\pi \cr} $$
