Đáp án:
$ x = - 40^{0} + k180^{0} $
$ x = - 25^{0} ± \dfrac{1}{2}arccos(- \dfrac{\sqrt{3}}{4}) + k.180^{0}$
Giải thích các bước giải:
Để cho gọn đặt $: t = 40^{0} + x ⇒ 3x = 3t - 120^{0}$
Ta có$: sin(80^{0} - x) = cos[90^{0} - (80^{0} - x)] = cos(10^{0} + x)$
$ ⇒ VT = \dfrac{1}{2}sin(10^{0} + x) - \dfrac{\sqrt{3}}{2}sin(80^{0} - x)$
$ = sin30^{0}sin(10^{0} + x) - cos30^{0}cos(10^{0} + x)$
$ = - cos(40^{0} + x) = - cost$
$ ⇒ VP = 2cos3x = 2cos(3t - 120^{0}) $
$ = 2(cos3t.cos120^{0} + sin3tsin120^{0}) = - cos3t + \sqrt{3}sin3t$
Thay vào $ PT ⇔ cos3t - cost - \sqrt{3}sin3t = 0$
$ ⇔ - 2sin2tsint - \sqrt{3}(3sint - 4sin³t) = 0$
$ ⇔ sint(4\sqrt{3}sin²t - 2sin2t - 3\sqrt{3}) = 0$
@ $ sint = 0 ⇔ t = k180^{0} ⇔ 40^{0} + x = k180^{0} $
$ ⇔ x = - 40^{0} + k180^{0} $
@ $ 4\sqrt{3}sin²t - 2sin2t - 3\sqrt{3} = 0$
$ ⇔ 2\sqrt{3}(1 - cos2t) - 2sin2t - 3\sqrt{3} = 0$
$ ⇔ \dfrac{\sqrt{3}}{2}cos2t + \dfrac{1}{2}sin2t = - \dfrac{\sqrt{3}}{4}$
$ ⇔ cos(2t - 30^{0}) = - \dfrac{\sqrt{3}}{4}$
$ ⇔ 2t - 30^{0} = ± arccos(- \dfrac{\sqrt{3}}{4}) + k.360^{0}$
$ ⇔ t = 15^{0} ± \dfrac{1}{2}arccos(- \dfrac{\sqrt{3}}{4}) + k.180^{0}$
$ ⇔ 40^{0} + x = 15^{0} ± \dfrac{1}{2}arccos(- \dfrac{\sqrt{3}}{4}) + k.180^{0}$
$ ⇔ x = - 25^{0} ± \dfrac{1}{2}arccos(- \dfrac{\sqrt{3}}{4}) + k.180^{0}$
