Đáp án:
$ x = \frac{π}{12} + (2k + 1)\frac{π}{4}$
$ x = - \frac{π}{6} + (2k + 1)\frac{π}{2} $
Giải thích các bước giải:
$PT ⇔ cos3x + \frac{\sqrt[]{3}}{2}sinx + \frac{1}{2}cosx = 0$
$ ⇔ cos3x + (cos\frac{π}{3}cosx + sin\frac{π}{3}sinx) = 0$
$ ⇔ cos3x + cos(x - \frac{π}{3}) = 0$
$ ⇔ 2cos(2x - \frac{π}{6})cos(x + \frac{π}{6}) = 0$
@ $ cos(2x - \frac{π}{6}) = 0 ⇔ 2x - \frac{π}{6} = (2k + 1)\frac{π}{2}$
$ ⇔ 2x = \frac{π}{6} + (2k + 1)\frac{π}{2} ⇔ x = \frac{π}{12} + (2k + 1)\frac{π}{4}$
@ $ cos(x + \frac{π}{6}) = 0 ⇔ x + \frac{π}{6} = (2k + 1)\frac{π}{2}$
$ ⇔ x = - \frac{π}{6} + (2k + 1)\frac{π}{2} $
