[TEX]\frac{1}{2}sin3x+\frac{\sqrt{3}}{2}.cos3x+\frac{1}{2}sin2x+\frac{\sqrt{3}}{2}.cos2x-(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx)=0[/TEX]
[TEX]sin(\frac{\pi }{6}).sin3x+cos(\frac{\pi }{6}).cos3x+sin(\frac{\pi }{6}).sin2x+cos(\frac{\pi }{6}).cos2x-[sin(\frac{\pi }{6}).sinx+cos(\frac{\pi }{6}).cosx)=0[/TEX]
[TEX] cos(3x-\frac{\pi }{6})+cos(2x-\frac{\pi }{6})-cos(x-\frac{\pi }{6})=0[/TEX]
[TEX]cos(3x-\frac{\pi }{6})=-sin(3x-\frac{2\pi}{3})=-2sin(\frac{3x}{2}-\frac{\pi}{3}).cos(\frac{3x}{2}-\frac{\pi}{3})=-2sin(\frac{3x}{2}-\frac{\pi}{3}).cos(\frac{3x}{2}-\frac{2\pi}{3})[/TEX]
Mà [TEX]cos(2x-\frac{\pi }{6})-cos(x-\frac{\pi }{6})=-2sin(\frac{3x}{2}-\frac{\pi}{6}).sin(\frac{x}{2})=-2cos(\frac{3x}{2}-\frac{2\pi}{3}).sin(\frac{x}{2})[/TEX]
Có nhân tử rùi đó.
Bạn này,
[TEX]cos(3x-\frac{\pi }{6})=-sin(3x-\frac{2\pi}{3})=-2sin(\frac{3x}{2}-\frac{\pi}{3}).cos(\frac{3x}{2}-\frac{\pi}{3})=-2sin(\frac{3x}{2}-\frac{\pi}{3}).cos(\frac{3x}{2}-\frac{2\pi}{3})[/TEX]
Mình không hiểu tại sao ở dấu bằng thứ 2 : [TEX]cos(\frac{3x}{2}-\frac{\pi}{3})[/TEX]
Lại có thể suy ra :[TEX]cos(\frac{3x}{2}-\frac{2\pi}{3})[/TEX]
Bạn giải thích giúp mình.