giải giúp mk với ạ đang cần gấp sin ^3 x+cos^3x=cosx với x thuộc [-pi,2pi]

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