Lời giải của Tự Học 365

$\begin{array}{l}\,\,\,\,\,\,3\sin 3x + \sqrt 3 \cos 9x = 1 + 4{\sin ^3}3x\\ \Leftrightarrow 3\sin 3x - 4{\sin ^3}3x + \sqrt 3 \cos 9x - 1 = 0\\ \Leftrightarrow \sin 9x + \sqrt 3 \cos 9x = 1\\ \Leftrightarrow \dfrac{1}{2}\sin 9x + \dfrac{{\sqrt 3 }}{2}\cos 9x = \dfrac{1}{2}\end{array}$

$\begin{array}{l} \Leftrightarrow \sin 9x\cos \dfrac{\pi }{3} + \cos 9x\sin \dfrac{\pi }{3} = \dfrac{1}{2}\\ \Leftrightarrow \sin \left( {9x + \dfrac{\pi }{3}} \right) = \sin \left( {\dfrac{\pi }{6}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}9x + \dfrac{\pi }{3} = \dfrac{\pi }{6} + k2\pi \\9x + \dfrac{\pi }{3} = \dfrac{{5\pi }}{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x =  - \dfrac{\pi }{{54}} + \dfrac{{k2\pi }}{9}\\x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{9}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\end{array}$

Đáp án cần chọn là: d