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