Đáp án:
Giải thích các bước giải:
$ PT ⇔ 4cos²x + 4cos²2x + 4cos²(\dfrac{π}{3} - 3x) = 7$
$ ⇔ 2(1 + cos2x) + 2(1 + cos4x) + 2[1 + cos(\dfrac{2π}{3} - 6x)] = 7$
$ ⇔ 2(cos4x + cos2x) + 2(cos\dfrac{2π}{3}cos6x + sin\dfrac{2π}{3}sin6x) = 1$
$ ⇔ 4cos3xcosx - (1 + cos6x) + \sqrt{3}sin6x = 0$
$ ⇔ 4cos3xcosx - 2cos²3x + 2\sqrt{3}sin3xcos3x = 0$
$ ⇔ 2cos3x(2cosx - cos3x + \sqrt{3}sin3x) = 0$
TH1 $: cos3x = 0 ⇔ 3x = \dfrac{π}{2} + kπ ⇔ x = \dfrac{π}{6} + k\dfrac{π}{3}$
TH2 $: 2cosx - cos3x + \sqrt{3}sin3x = 0 $
$ ⇔ \dfrac{1}{2}cos3x - \dfrac{\sqrt{3}}{2}sin3x = cosx$
$ ⇔ cos(3x + \dfrac{π}{3}) = cosx$
$ ⇔ 3x + \dfrac{π}{3} = ± x + k2π$
$ ⇔ x = - \dfrac{π}{6} + kπ; x = - \dfrac{π}{12} + k\dfrac{π}{2} $
