[TEX]\sqrt{2}sin(\frac{2x}{3} -\frac{\pi}{3}) -\sqrt{6}sin(\frac{2x}{3} +\frac{\pi}{6})=2sin(\frac{3x}{2} -\frac{\pi}{6})-2cos(\frac{x}{6}+\frac{2\pi}{3}[/TEX]
Ta để ý rằng :
[TEX]sin(\frac{2x}{3} +\frac{\pi}{6}) = cos ( \frac{2x}{3} - \frac{\pi}{3})[/TEX] nên phương trình đã cho tương đương với :
[TEX]2\sqrt{2} sin {\(\frac{2x}{3} - \frac{2\pi}{3}\) } = 2\left( ( sin{ \( {\frac{3x}{2} - \frac{\pi}{6}} \) }- sin{ \( - \frac{\pi}{6} - \frac{x}{6} \) }\right)[/TEX]
[TEX]\Rightarrow \sqrt{2} sin {\(\frac{2x}{3} - \frac{2\pi}{3}\)= 2cos {\( \frac{2x}{3} - \frac{\pi}{6}\)} . sin (\frac{5x}{6} ) \ \ \ \ \ \ \ \ \ \ \ \ \ (1)[/TEX]
Lại để ý rằng : [TEX]sin ( \frac{2x}{3} - \frac{2 \pi}{3} ) = cos ( \frac{7 \pi}{6} - \frac{2x}{3}) = - cos ( \frac{2x}{3} - \frac{\pi}{6})[/TEX]
nên ta có :
[TEX](1) \Leftrightarrow \left[cos ( \frac{2x}{3} - \frac{\pi}{6}) = 0 \\ sin (\frac{5x}{6} ) = \frac{- \sqrt{2}}{2}[/TEX]
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