Đáp án:
Giải thích các bước giải:
$\sqrt{3}sin3x+cos3x=\sqrt{2}\Rightarrow \frac{\sqrt{3}}{2}sin3x+\frac{1}{2}cos3x=\frac{\sqrt{2}}{2}\Rightarrow cos\frac{\pi}{6}sin3x+sin\frac{\pi}{3}cos3x=sin\frac{\pi}{4}\Leftrightarrow sin(\frac{\pi}{3}-3x)=sin\frac{\pi}{4}\Rightarrow \begin{bmatrix}
\frac{\pi}{3}-3x= \frac{\pi}{4}+k2\pi& & \\
\frac{\pi}{3}-3x=\frac{3\pi}{4}+k2\pi & &
\end{bmatrix}\Rightarrow x=...$
