Đáp án:
$ x = ± \dfrac{π}{3} + kπ$
$ x = \dfrac{π}{10} + k\dfrac{2π}{5}$
$ x = \dfrac{π}{2} + k2π$
Giải thích các bước giải:
$ PT ⇔ (sin5x + sinx) + sin3x - (cos4x + 1) - cos2x = 0$
$ ⇔ 2sin3xcos2x + sin3x - 2cos²2x - cos2x = 0$
$ ⇔ sin3x(2cos2x + 1) - cos2x(2cos2x + 1) = 0$
$ ⇔ (2cos2x + 1)(sin3x - cos2x) = 0$
TH1 $: 2cos2x + 1 = 0 ⇔ cos2x = - \dfrac{1}{2}$
$ ⇔ 2x = ± \dfrac{2π}{3} + k2π ⇔ x = ± \dfrac{π}{3} + kπ$
TH2 $: sin3x - cos2x = 0 ⇔ sin3x = cos2x = sin(\dfrac{π}{2} - 2x)$
$3x = \dfrac{π}{2} - 2x + k2π ⇔ x = \dfrac{π}{10} + k\dfrac{2π}{5}$
$3x = π - (\dfrac{π}{2} - 2x) + k2π ⇔ x = \dfrac{π}{2} + k2π$
