Đáp án: $x = \dfrac{\pi }{4} + k\dfrac{\pi }{2}\left( {k \in Z} \right)$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
{\sin ^3}x + {\cos ^3}x = 2\left( {{{\sin }^5}x + {{\cos }^5}x} \right)\\
\Leftrightarrow {\sin ^3}x\left( {2{{\sin }^2}x - 1} \right) + {\cos ^3}x\left( {2{{\cos }^2}x - 1} \right) = 0\\
\Leftrightarrow {\sin ^3}x.\left( { - \cos 2x} \right) + {\cos ^3}x.\cos 2x = 0\\
\Leftrightarrow \cos 2x\left( {{{\sin }^3}x - {{\cos }^3}x} \right) = 0\\
\Leftrightarrow \cos 2x\left( {\sin x - \cos x} \right)\left( {{{\sin }^2}x + \sin x.\cos x + {{\cos }^2}x} \right) = 0\\
\Leftrightarrow \cos 2x\left( {\sin x - \cos x} \right)\left( {1 + \dfrac{{\sin 2x}}{2}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 2x = 0\\
\sin x - \cos x = 0\\
\sin 2x = - 2\left( {vl} \right)
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos 2x = 0\\
\cos \left( {x + \dfrac{\pi }{4}} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} + k\pi \\
x + \dfrac{\pi }{4} = \dfrac{\pi }{2} + k\pi
\end{array} \right.\left( {k \in Z} \right) \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\dfrac{\pi }{2}\\
x = \dfrac{\pi }{4} + k\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Leftrightarrow x = \dfrac{\pi }{4} + k\dfrac{\pi }{2}\left( {k \in Z} \right)
\end{array}$
Vậy phương trình có họ nghiệm là: $x = \dfrac{\pi }{4} + k\dfrac{\pi }{2}\left( {k \in Z} \right)$
