Đáp án: $ m = \dfrac{1 ± \sqrt{17}}{8}$
Giải thích các bước giải:
$PT ⇔ cos5x + cosx + cos5x + cos3x = 0$
$ ⇔ 2cos3xcos2x + 2cos4xcosx = 0$
$ ⇔ (4cos³x - 3cosx)cos2x + (2cos²2x - 1)cosx = 0$
$ ⇔ cosx[(4cos²x - 3)cos2x + (2cos²2x - 1)] = 0$
$ ⇔ cosx[(2cos2x - 1)cos2x + (2cos²2x - 1)] = 0$
$ ⇔ cosx(4cos²2x - cos2x - 1) = 0$
@ $ cosx = 0 ⇔ x = \dfrac{π}{2} + kπ$
@ $ 4cos²2x - cos2x - 1 = 0 ⇒ cos2x = \dfrac{1 ± \sqrt{17}}{8}$
$ ⇔ 2x = ± arccos\dfrac{1 ± \sqrt{17}}{8} + k2π$
$ ⇔ x = ± \dfrac{1}{2} arccos\dfrac{1 ± \sqrt{17}}{8} + kπ$
$ ⇒ m = \dfrac{1 ± \sqrt{17}}{8}$
