Các bạn giải nha ! Thử sức xem có ăn 1 điểm trong đề thi ĐH hem !
[TEX]\begin{array}{l} (\cos 2x - \cos 4x)^2 - 4 = 2(1 + sin3x) \\ \Leftrightarrow (\cos 2x - \cos 4x - 2)(\cos 2x - \cos 4x + 2) = - 2(4\sin ^3 x - 3sinx- 1) \\ \Leftrightarrow (\cos 2x - 1 - 2\cos ^2 2x)(\cos 2x + 1 + 2\sin ^2 2x) = 2(1 - sinx)(2\sin x + 1)^2 (*) \\ \end{array}[/TEX]
Ta có:
[TEX]\begin{array}{l} i)\cos 2x - 1 - \cos ^2 2x < 0 \\ \cos 2x + 1 + 2\sin ^2 2x \ge 0 \\ \Rightarrow (\cos2x - 1 - 2\cos ^2 2x)(\cos 2x + 1 + 2\sin ^2 2x) \le 0 \\ \end{array}[/TEX]
[TEX]ii)2(1 - \sin x)(2\sin x + 1)^2 \ge 0[/TEX]
Do đó
[TEX](*) \Leftrightarrow \left\{ \begin{array}{l} \cos 2x = - 1 \\ \sin 2x = 0 \\ \left[ \begin{array}{l} \sin x = - \frac{1}{2} \\ \sin x = 1 \\ \end{array} \right. \\ \end{array} \right.[/TEX]
[TEX]\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} \cos 2x = - 1 \\ \sin x = - \frac{1}{2} \\ \end{array} \right. \\ \left\{ \begin{array}{l} \cos 2x = - 1 \\ \sin x = 1 \\ \end{array} \right. \\ \end{array} \right. \\ \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} \cos x = 0 \\ \sin x = - \frac{1}{2} \\ \end{array} \right. \\ \left\{ \begin{array}{l} \cos x = 0 \\ \sin x = 1 \\ \end{array} \right. \\ \end{array} \right. \\ \Leftrightarrow x = \frac{\pi }{2} + k2\pi (k \in Z). \\ \end{array}[/TEX]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
