Giải thích các bước giải:
\(\begin{array}{l}y = 2(1 + \sin 2x\cos 4x) - \frac{1}{2}(\cos 4x - \cos 8x)\\ = 2 + \sin 6x - \sin 2x - \sin 6x\sin 2x\\ = 1 + (1 - \sin 2x) + \sin 6x(1 - \sin 2x)\\ = 1 + (1 - \sin 2x)(1 + \sin 6x)\end{array}\)
Ta cos : \(\left\{ \begin{array}{l}\sin 2x \ge - 1\\\sin 6x \le 1\end{array} \right. \Rightarrow y \le 1 + 2.2 = 5\)
Dấu “=” xảy ra với \(\left\{ \begin{array}{l}\sin 2x = - 1\\\sin 6x = 1\end{array} \right. \Rightarrow x = \frac{{3\pi }}{4}\)
\(\left\{ \begin{array}{l}\sin 2x \le 1\\\sin 6x \ge - 1\end{array} \right. \Rightarrow y \ge 1\)
Dấu “=” xảy ra với \(\left\{ \begin{array}{l}\sin 2x = 1\\\sin 6x = - 1\end{array} \right. \Rightarrow x = \frac{\pi }{4}\)
Vậy \(\max \,\,y = 5 & & ;\,\,\,\min \,\,y = 1.\)
