Đáp án:

Giải thích các bước giải:

g) $PT ⇔ 3cos²x - sin²x - 2sinxcosx = 0$

$ ⇔ (cosx - sinx)(3cosx + sinx) = 0$

TH1 $: cosx - sinx = 0 ⇔ tanx = 1 ⇔ x = \dfrac{π}{4} + kπ$

TH2 $: 3cosx + sinx = 0 ⇔ tanx = - 3 ⇔ x = arctan(-3) + kπ$

h) $ PT ⇔\sqrt{13}sin4x = 2sin2x - 3cos2x $

$ ⇔ sin4x = \dfrac{2}{\sqrt{13}}sin2x - \dfrac{3}{\sqrt{13}}cos2x $

$ ⇔ sin4x = sin(2x - a) (a = arccos\dfrac{2}{\sqrt{13}} = arcsin\dfrac{3}{\sqrt{13}})$

TH1 $: 4x = 2x - a + k2π $

$ ⇔ x = - \dfrac{a}{2} + kπ = - \dfrac{1}{2}arccos\dfrac{2}{\sqrt{13}} + kπ $

TH2 $: 4x = π - (2x - a) + k2π$

$ ⇔ x = \dfrac{a}{6} + (2k + 1)\dfrac{π}{6} = \dfrac{1}{6}arccos\dfrac{2}{\sqrt{13}} + (2k + 1)\dfrac{π}{6} $

i)Để ý $: - sin2x = - 2sinxcosx = (cosx - sinx)² - 1$

$PT ⇔ 2(cos²x - sin²x) - 2(sinx + cosx) + (cosx - sinx)² - 1 = 0$

$ ⇔ 2(sinx + cosx)(cosx - sinx - 1) + (cosx - sinx + 1)(cosx - sinx - 1) = 0$

$ ⇔ (cosx - sinx - 1)(3cosx + sinx + 1) = 0$

TH1 $: cosx - sinx - 1 = 0 ⇔ cosx - sinx = 1$

$ ⇔ \dfrac{\sqrt{2}}{2}cosx - \dfrac{\sqrt{2}}{2}sinx = \dfrac{\sqrt{2}}{2}$

$ ⇔ cos(x + \dfrac{π}{4}) = cos\dfrac{π}{4} $

$ ⇔ x + \dfrac{π}{4} = ± \dfrac{π}{4} + k2π $

$ ⇔ x = k2π; x = - \dfrac{π}{2} + k2π $

TH2 $: 3cosx + sinx + 1 = 0 ⇔ 3cosx + sinx = - 1$

$ ⇔ \dfrac{3}{\sqrt{10}}cosx + \dfrac{1}{\sqrt{10}}sinx = - \dfrac{1}{\sqrt{10}}$

$ ⇔ cos(x - a) = - \dfrac{1}{\sqrt{10}} (a = arccos\dfrac{3}{\sqrt{10}} = arcsin\dfrac{1}{\sqrt{10}}) $

$ ⇔ x - a = ± arccos(-\dfrac{1}{\sqrt{10}}) + k2π $

$ ⇔ x = a ± arccos(-\dfrac{1}{\sqrt{10}}) + k2π $

$ = arccos\dfrac{3}{\sqrt{10}} ± arccos(-\dfrac{1}{\sqrt{10}}) + k2π $

j) $PT ⇔ sin17x = - \dfrac{1}{2}sin5x - \dfrac{\sqrt{3}}{2}cos5x $

$ ⇔ sin17x = sin5xcos\dfrac{2π}{3} - sin\dfrac{2π}{3}cos5x $

$ ⇔ sin17x = sin(5x - \dfrac{2π}{3})$

TH1 $: 17x = 5x - \dfrac{2π}{3} + k2π ⇔ x = - \dfrac{π}{18} + k\dfrac{π}{6}$

TH2 $: 17x = π - (5x - \dfrac{2π}{3}) + k2π ⇔ x = \dfrac{π}{33} + (2k + 1)\dfrac{π}{22}$