Đáp án:

Giải thích các bước giải:

a) $ 4(sin^{4}\frac{x}{2} + cos^{4}\frac{x}{2}) + \sqrt[]{3}sin2x = 2$

$ ⇔ 4[(sin²\frac{x}{2} + cos²\frac{x}{2})² - 2sin²\frac{x}{2}cos²\frac{x}{2}] + \sqrt[]{3}sin2x = 2$

$ ⇔ 4 - 2(2sin\frac{x}{2}cos\frac{x}{2})² + \sqrt[]{3}sin2x = 2$

$ ⇔ 2(1 - sin²x) + \sqrt[]{3}sin2x = 0$ 

$ ⇔ 2cos²x + 2\sqrt[]{3}sinxcosx = 0$

$ ⇔ 4cosx(\frac{1}{2}cosx + \frac{\sqrt[]{3}}{2}sinx) = 0$  

$ ⇔ 4cosxcos(x - \frac{π}{3}) = 0$  

@ $cosx = 0 ⇔ x = (2k + 1)\frac{π}{2}$

@ $cos(x - \frac{π}{3}) = 0 ⇔ x - \frac{π}{3} = (2k + 1)\frac{π}{2} ⇔ x = \frac{π}{3} + (2k + 1)\frac{π}{2}$

b) $ 1 + sin³x + cos³x = \frac{3}{2}sin2x$

$ ⇔ 1 + (sinx + cosx)³ - 3sinxcosx(sin²x + cos²x) = \frac{3}{2}sin2x$

$ ⇔ 1 + (sinx + cosx)³ - 3sin2x = 0$

Đặt $ t = sinx + cosx = \sqrt[]{2}sin(x + \frac{π}{4}) ⇒ |t| ≤ \sqrt[]{2}$ 

$t² = sin²x + cos²x + 2sinxcosx = 1 + sin2x ⇔ sin2x = t² - 1$ thay vào $PT$

$ 1 + t³ - 3(t² - 1) ⇔ t³ - 3t² + 4 = 0 ⇔ (t + 1)(t - 2)² = 0$

$ ⇔ t + 1 = 0 ⇔ t = - 1$ ( vì $|t| ≤ \sqrt[]{2} ⇒ t - 2 < 0)$

$ ⇔ \sqrt[]{2}sin(x + \frac{π}{4}) = - 1 ⇔ sin(x + \frac{π}{4}) = - \frac{\sqrt[]{2}}{2}$

@ $ x + \frac{π}{4} = - \frac{π}{4} + k2π ⇔ x = - \frac{π}{2} + k2π$

@ $ x + \frac{π}{4} = - \frac{3π}{4} + k2π ⇔ x = (2k - 1)π$

c) $cosx + cos2x + cos3x + cos4x = 0$

$ ⇔ (cosx + cos3x) + (cos2x + cos4x) = 0$

$ ⇔ 2cos2xcosx + 2cos3xcosx = 0$

$ ⇔ 2cosx(cos2x + cos3x) = 0$

$ ⇔ 4cosxcos\frac{5x}{2}cos\frac{x}{2} = 0$

@ $ cosx = 0 ⇔ x = (2k + 1)\frac{π}{2} $

@ $ cos\frac{5x}{2} = 0 ⇔ \frac{5x}{2} = (2k + 1)\frac{π}{2} ⇔ x = (2k + 1)\frac{π}{5}$

@ $ cos\frac{x}{2} = 0 ⇔ \frac{x}{2} = (2k + 1)\frac{π}{2} ⇔ x = (2k + 1)π$