Giải pt$ a, \cos2x + \sqrt{3} \sin2x = 2 \cos3x$$b, \sin2x - \cos 2x = \sqrt{2} \sin(x+\pi/6)$$c, \sin^2x - \sin2x - 3\cos^2x =0$$d, 3\sin^2x - \sin 2x + \cos^2x = 2$Bài 2 : tìm max min của h.s$a, y = 2\cos2x - \sin x$$b, y= 3\cos2x - 2\sin2x$$c, y = \sin^2x - \sin2x + 3cos^2x$