`~rai~`
\(\sqrt{3}\sin3x+\cos3x=1\\\Leftrightarrow \dfrac{\sqrt{3}}{2}\sin3x+\dfrac{1}{2}\cos3x=\dfrac{1}{2}\\\Leftrightarrow \cos\dfrac{\pi}{6}\sin3x+\sin\dfrac{\pi}{6}\cos3x=\dfrac{1}{2}\\\Leftrightarrow \sin\left(3x+\dfrac{\pi}{6}\right)=\dfrac{1}{2}.\\\to\text{Chọn ý D.}\)