Đáp án:
$ x = \dfrac{5π}{18} + 2k\frac{π}{3}$
$ x = \dfrac{5π}{6} + (2k+ 1)π $
Giải thích các bước giải:
ĐKXĐ $: 1 - 2sinx \neq 0 ⇔ sinx \neq \dfrac{1}{2} $
$ ⇔ x \neq \dfrac{π}{6} + k2π; x \neq \dfrac{5π}{6} + k2π; (1)$
Ta có $: 4cos²\dfrac{x}{2} = 2(cosx + 1); cos(2x - 3π) = - cos2x$
$ 2cos²(\dfrac{7π}{4} - x) = cos(\dfrac{7π}{2} - 2x) + 1 $
$ = cos(3π + \dfrac{π}{2} - 2x) + 1 = - cos(\dfrac{π}{2} - 2x) + 1 = - sin2x + 1$
Thay vào $PT:$
$ 2(cosx + 1) - sin2x + 1 + \sqrt[]{3}cos2x - 3 = 0$
$ ⇔ 2cosx - sin2x + \sqrt[]{3}cos2x = 0$
$ ⇔ cosx + \dfrac{\sqrt[]{3}}{2}cos2x - \dfrac{1}{2}sin2x = 0 $
$ ⇔ cosx + cos(2x + \dfrac{π}{6}) = 0 $
$ ⇔ 2cos(\dfrac{3x}{2} + \dfrac{π}{12})cos(\dfrac{x}{2} + \dfrac{π}{12}) =0$
@ $ cos(\dfrac{3x}{2} + \dfrac{π}{12}) = 0 ⇔ \dfrac{3x}{2} + \dfrac{π}{12} = (2k + 1)\dfrac{π}{2}$
$ ⇔ \dfrac{3x}{2} = - \dfrac{π}{12} + (2k + 1)\dfrac{π}{2} ⇔ x = \dfrac{5π}{18} + 2k\dfrac{π}{3} (TM (1))$
@ $ cos(\dfrac{x}{2} + \dfrac{π}{12}) = 0 ⇔ \dfrac{x}{2} + \dfrac{π}{12} = (2k + 1)\dfrac{π}{2}$
$ ⇔ \dfrac{x}{2} = - \dfrac{π}{12} + (2k + 1)\dfrac{π}{2} ⇔ x = \dfrac{5π}{6} + kπ $
Loại họ nghiệm $: x = \dfrac{5π}{6} + k2π (ko TM (1))$
Chọn họ nghiệm $: x = \dfrac{5π}{6} + (2k+ 1)π (TM (1))$
