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