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Question

Giải phương trình: $\sqrt{2}\left(\sin x+\sqrt{3}\cos x\right)=\sqrt{3}\cos 2x-\sin 2x$

cobengoc59 · Answer 1 · 4/20/2013 2:28:02 PM

ta có :

$2\sqrt{2}(\frac{1}{2}\sin x+\frac{\sqrt{3}}{2}\cos x)=2(\frac{\sqrt{3}}{2}\cos 2x-\frac{1}{2}\sin 2x$

$\Leftrightarrow 2\sqrt{2}(\cos \frac{pi}{3}\sin x+\sin \frac{pi}{3}\cos x)=2(\sin \frac{pi}{3}\cos 2x-\cos \frac{pi}{3}\sin 2x)$

$\Leftrightarrow 2\sqrt{2}(\sin (\frac{pi}{3}+x)=2\sin (\frac{pi}{3}-2x)$

$\Leftrightarrow \sqrt{2}\cos (\frac{pi}{2}-\frac{pi}{3}-x)=\sin (2(\frac{pi}{6}-x))$

$\Leftrightarrow \sqrt{2}\cos (\frac{pi}{6}-x)=2sin(\frac{pi}{6}-x)\cos (\frac{pi}{6}-x)$

TH1:

$\cos(\frac{pi}{6}-x)=0$

$\Leftrightarrow x=-\frac{pi}{3}-kpi$

TH2:

$\sin (\frac{pi}{6}-x)=\frac{1}{\sqrt{2}}$

$\Leftrightarrow \sin (\frac{pi}{6}-x)=sin\frac{pi}{4}$

$\Leftrightarrow x=-\frac{1}{12}-k2pi hoặc x=\frac{-7}{12}-k2pi$