Đáp án:
\[D\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\sin ^3}x - \sqrt 3 {\cos ^3}x = \sin x.{\cos ^2}x - \sqrt 3 {\sin ^2}x.\cos x\\
\Leftrightarrow {\sin ^3}x - \sqrt 3 {\cos ^3}x - \sin x.{\cos ^2}x + \sqrt 3 {\sin ^2}x.\cos x = 0\\
\Leftrightarrow \left( {{{\sin }^3}x + \sqrt 3 {{\sin }^2}x.\cos x} \right) - \left( {\sqrt 3 {{\cos }^3}x + \sin x.{{\cos }^2}x} \right) = 0\\
\Leftrightarrow {\sin ^2}x.\left( {\sin x + \sqrt 3 \cos x} \right) - {\cos ^2}x.\left( {\sqrt 3 \cos x + \sin x} \right) = 0\\
\Leftrightarrow \left( {\sin x + \sqrt 3 \cos x} \right)\left( {{{\sin }^2}x - {{\cos }^2}x} \right) = 0\\
\Leftrightarrow 2.\left( {\dfrac{1}{2}\sin x + \dfrac{{\sqrt 3 }}{2}\cos x} \right).\left[ { - \left( {{{\cos }^2}x - {{\sin }^2}x} \right)} \right] = 0\\
\Leftrightarrow 2.\left( {\sin x.\cos \dfrac{\pi }{3} + \cos x.\sin \dfrac{\pi }{3}} \right).\left( { - \cos 2x} \right) = 0\\
\Leftrightarrow 2\sin \left( {x + \dfrac{\pi }{3}} \right).\cos 2x = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin \left( {x + \dfrac{\pi }{3}} \right) = 0\\
\cos 2x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{3} = k\pi \\
2x = \dfrac{\pi }{2} + k\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{\pi }{3} + k\pi \\
x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
Vậy đáp án đúng là \(D\)
