Đáp án:
1) $ x = \dfrac{π}{3} + kπ$
2) $ x = - \dfrac{5π}{6} + k2π; x = - \frac{π}{18} + k\dfrac{2π}{3}$
Giải thích các bước giải: Tham khảo
1) $ PT ⇔ (1 - cos2x) + \sqrt{3}sin2x = 3$
$ ⇔ \sqrt{3}sin2x - cos2x = 2$
$ ⇔ \dfrac{\sqrt{3}}{2}sin2x - \dfrac{1}{2}cos2x = 1$
$ ⇔ sin2xcos\dfrac{π}{6} - cos2xsin\dfrac{π}{6} = 1$
$ ⇔ sin(2x - \dfrac{π}{6}) = sin\dfrac{π}{2}$
$ ⇔ 2x - \dfrac{π}{6} = \dfrac{π}{2} + k2π ⇔ x = \dfrac{π}{3} + kπ$
2) $ cos(x + \dfrac{π}{2}) = - sinx; sin(x - \dfrac{π}{2}) = - cosx$
Thay vào $PT ⇔ - \sqrt{3}sinx - cosx = 2sin2x$
$ ⇔ \dfrac{\sqrt{3}}{2}sinx + \dfrac{1}{2}cosx = - sin2x$
$ ⇔ sinxcos\dfrac{π}{6} + cosxsin\dfrac{π}{6} = sin(2x + π) $
$ ⇔ sin(x + \dfrac{π}{6}) = sin(2x + π) $
@ $ 2x + π = x + \dfrac{π}{6} + k2π ⇔ x = - \dfrac{5π}{6} + k2π$
@ $ 2x + π = π - (x + \dfrac{π}{6}) + k2π ⇔ x = - \dfrac{π}{18} + k\dfrac{2π}{3}$
