Đáp án:
\(\left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \arctan \left( { - 2} \right) + k\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\sin ^3}x - 3\sin x.{\cos ^2}x + 2{\cos ^3}x = 0\,\,\,\,\,\,\left( 1 \right)\\
\Leftrightarrow \left( {{{\sin }^3}x - \sin x.{{\cos }^2}x} \right) + \left( { - 2\sin x.{{\cos }^2}x + 2{{\cos }^3}x} \right) = 0\\
\Leftrightarrow \sin x.\left( {{{\sin }^2}x - {{\cos }^2}x} \right) - 2{\cos ^2}x.\left( {\sin x - \cos x} \right) = 0\\
\Leftrightarrow \sin x.\left( {\sin x - \cos x} \right)\left( {\sin x + \cos x} \right) - 2{\cos ^2}x.\left( {\sin x - \cos x} \right) = 0\\
\Leftrightarrow \left( {\sin x - \cos x} \right).\left[ {\sin x.\left( {\sin x + \cos x} \right) - 2{{\cos }^2}x} \right] = 0\\
\Leftrightarrow \left( {\sin x - \cos x} \right).\left( {{{\sin }^2}x + \sin x.\cos x - 2{{\cos }^2}x} \right) = 0\\
\Leftrightarrow \left( {\sin x - \cos x} \right).\left[ {\left( {{{\sin }^2}x - \sin x.\cos x} \right) + \left( {2\sin x.\cos x - 2{{\cos }^2}x} \right)} \right] = 0\\
\Leftrightarrow \left( {\sin x - \cos x} \right).\left[ {\sin x.\left( {\sin x - \cos x} \right) + 2\cos x.\left( {\sin x - \cos x} \right)} \right] = 0\\
\Leftrightarrow \left( {\sin x - \cos x} \right).\left( {\sin x - \cos x} \right).\left( {\sin x + 2\cos x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x - \cos x = 0\\
\sin x + 2\cos x = 0
\end{array} \right.\,\,\,\,\,\left( * \right)\\
TH1:\,\,\,\,\cos x = 0\\
\left( 1 \right) \Leftrightarrow {\sin ^3}x = 0 \Leftrightarrow \sin x = 0\\
\Rightarrow {\sin ^2}x + {\cos ^2}x = 0\,\,\,\,\,\left( L \right)\\
TH2:\,\,\,\cos x \ne 0\\
\left( * \right) \Leftrightarrow \left[ \begin{array}{l}
\dfrac{{\sin x - \cos x}}{{\cos x}} = 0\\
\dfrac{{\sin x + 2\cos x}}{{\cos x}} = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{{\sin x}}{{\cos x}} - 1 = 0\\
\dfrac{{\sin x}}{{\cos x}} + 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\tan x = 1\\
\tan x = - 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \arctan \left( { - 2} \right) + k\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
