Giải thích các bước giải:
$\sqrt[]{3}cos2x +sin2x + 2sin(2x-\frac{\pi}{6})=2\sqrt{2}$
=> $\frac{\sqrt{3}}{2}cos2x + \frac{1}{2}sin2x + sin(2x-\frac{\pi}{6}) = \sqrt{2}$
=> $cos\frac{\pi}{6}.cos2x+sin\frac{\pi}{6}.sin2x + sin(2x-\frac{\pi}{6}) = \sqrt{2} $
=> $cos(2x-\frac{\pi}{6}) + sin(2x-\frac{\pi}{6}) = \sqrt{2}$
=> $\frac{\sqrt{2}}{2}.cos(2x-\frac{\pi}{6}) + \frac{\sqrt{2}}{2}.sin(2x-\frac{\pi}{6}) = 1$
=> $sin\frac{\pi}{4}.cos(2x-\frac{\pi}{6}) + cos\frac{\pi}{4}.sin(2x-\frac{\pi}{6})=1$
=> $sin(2x+\frac{\pi}{12})=1$
=> $x = \frac{5\pi}{24} + k\pi$
