Đáp án:

$ x = - \frac{π}{16} = (2k + 1)\frac{π}{8}$

$ x = arctan(2 - \sqrt[]{3}) + kπ$

Giải thích các bước giải:

$ cos5x - sin3x = \sqrt[]{3}(cos3x - sin5x)$ 

$ ⇔ cos5x + cos(3x + \frac{π}{2}) = \sqrt[]{3}[cos3x + cos(5x + \frac{π}{2})]$ 

$ ⇔ 2cos(4x + \frac{π}{4})cos(x - \frac{π}{4}) - 2\sqrt[]{3}cos(4x + \frac{π}{4})cos(x + \frac{π}{4}) = 0$ 

$ ⇔ 2cos(4x + \frac{π}{4})[cos(x - \frac{π}{4}) - \sqrt[]{3}cos(x + \frac{π}{4})] = 0$ 

@ $ cos(4x + \frac{π}{4}) = 0 ⇔ 4x + \frac{π}{4} = (2k + 1)\frac{π}{2}$

$ ⇔ x = - \frac{π}{16} + (2k + 1)\frac{π}{8}$

@ $ cos(x - \frac{π}{4}) - \sqrt[]{3}cos(x + \frac{π}{4}) = 0$

$ ⇔ \sqrt[]{2}(cosx + sinx) - \sqrt[]{6}(cosx - sinx) = 0$

$ ⇔ (\sqrt[]{3} + 1)sinx = (\sqrt[]{3} - 1)cosx$

$ ⇔ tanx = \frac{\sqrt[]{3} - 1}{\sqrt[]{3} + 1} = 2 - \sqrt[]{3}$ 

$ ⇒ x = arctan(2 - \sqrt[]{3}) + kπ$