Đáp án:
\(x \in \left\{ { - \frac{{3\pi }}{4};\,\,\frac{\pi }{4};\,\,\frac{{5\pi }}{4};\,\, - \pi ;\,\,0;\,\,\pi ;\,\,2\pi } \right\}\)
Giải thích các bước giải:
\(\begin{array}{l}
{\sin ^3}x + {\cos ^3}x = \cos x\,\,\,\left( * \right)\\
Xet\,\,\,\cos x = 0 \Rightarrow \left( * \right) \Leftrightarrow {\sin ^3}x = 0\,\,\,\left( {vo\,\,\,ly} \right)\\
\Rightarrow \cos x = 0\,\,\,khong\,\,\,la\,\,nghiem\,\,cua\,\,pt\,\,\,\left( * \right)\\
Chia\,\,\,ca\,\,\,2\,\,\,ve\,\,\,cua\,\,\left( * \right)\,\,\,cho\,\,\,{\cos ^3}x\,\,\,ta\,\,\,duoc:\\
\left( * \right) \Leftrightarrow {\tan ^3}x + 1 = \frac{1}{{{{\cos }^2}x}}\\
\Leftrightarrow {\tan ^3}x + 1 = 1 + {\tan ^2}x\\
\Leftrightarrow {\tan ^3}x - {\tan ^2}x = 0\\
\Leftrightarrow {\tan ^2}x\left( {\tan x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\tan x = 1\\
\tan x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{4} + k\pi \\
x = m\pi
\end{array} \right.\,\,\,\left( {k,\, m\in Z} \right).
\end{array}\)
\(\begin{array}{l}
x \in \left[ { - \pi ;\,\,2\pi } \right]\\
\Rightarrow \left[ \begin{array}{l}
- \pi \le \frac{\pi }{4} + k\pi \le 2\pi \\
- \pi \le m\pi \le 2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
- \frac{{5\pi }}{4} \le k\pi \le \frac{{7\pi }}{4}\\
- 1 \le m \le 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
- \frac{5}{4} \le k \le \frac{7}{4}\\
- 1 \le m \le 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
k \in \left\{ { - 1;\,\,0;\,\,1} \right\}\\
m \in \left\{ { - 1;\,\,0;\,\,1;\,\,2} \right\}
\end{array} \right.\\
\Rightarrow x \in \left\{ { - \frac{{3\pi }}{4};\,\,\frac{\pi }{4};\,\,\frac{{5\pi }}{4};\,\, - \pi ;\,\,0;\,\,\pi ;\,\,2\pi } \right\}.
\end{array}\)
