Giải phương trình cosx + cos3x + 2cos5x = 0 A. x=pi/2+kpi, x=+-arccos1+căn bậc hai của 17/8+kpi,

* Hướng dẫn giải

cosx + cos3x + 2cos5x = 0

⇔ cosx + cos3x + cos5x + cos5x = 0

⇔ (cosx + cos5x) + (cos3x + cos5x) = 0

⇔ 2cos3xcos2x + 2cos4xcosx = 0
⇔ 2(4cos³ x − 3cosx)cos2x + 2cos4xcosx = 0

⇔2cosx(4cos² x − 3)cos2x + 2cos4xcosx = 0

⇔ 2cosx[(4cos² x − 3)cos2x + cos4x] = 0

⇔ 2cosx[[2(1 + cos2x) − 3]cos2x + 2cos² 2x − 1] = 0

⇔ 2cosx[(2cos2x − 1)cos2x + 2cos² 2x − 1] = 0

⇔ 2cosx[4cos² 2x − cos2x − 1] = 0

$\iff \left[\begin{array}{c}\cos x=0\\ \cos 2x=\frac{1+\sqrt{17}}{8}\\ \cos 2x=\frac{1-\sqrt{17}}{8}\end{array}\right.$

$\iff \left[\begin{array}{c}x=\frac{\pi }{2}+k\pi \\ 2x=\pm \arccos \frac{1+\sqrt{17}}{8}+k2\pi \\ 2x=\pm \arccos \frac{1-\sqrt{17}}{8}+k2\pi \end{array}\right.$

$\iff \left[\begin{array}{c}x=\frac{\pi }{2}+k\pi \\ x=\pm \frac{1}{2}\arccos \frac{1+\sqrt{17}}{8}+k\pi \\ x=\pm \frac{1}{2}\arccos \frac{1-\sqrt{17}}{8}+k\pi \end{array}\right.$

Vậy nghiệm của phương trình là: $x=\frac{\pi }{2}+k\pi ,x=\pm \frac{1}{2}\arccos \frac{1+\sqrt{17}}{8}+k\pi ,x=\pm \frac{1}{2}\arccos \frac{1-\sqrt{17}}{8}+k\pi$

Đáp án cần chọn là: D