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Question

$(\cot x-1)(1-\sqrt 3 \cos 4x)=2\sin (2x-\dfrac{3\pi}{2})$

♂Vitamin_Tờ♫ · Answer 1 · 11/19/2013 6:57:09 PM

ĐK:

$\sin x\neq 0\Leftrightarrow x\neq k\pi,k\in Z$

$+\sin(2x-\frac{3\pi}{2})=\sin(2x+\frac{\pi}{2}-2\pi)=\sin(2x+\frac{\pi}{2})=\cos2x$

$pt\Leftrightarrow (\cos x-\sin x)(1-\sqrt{3}\cos4x)=2\cos2x\sin x$

$\Leftrightarrow (\cos x-\sin x)(1-\sqrt{3}\cos4x)=2\sin x(\cos x-\sin x)(\cos x+\sin x)$

$\Leftrightarrow (\cos x-\sin x)(1-\sqrt{3}\cos4x-\sin2x-2\sin^2x)=0$

$\Leftrightarrow (\cos x-\sin x)(\cos2x-\sin2x-\sqrt{3}(\cos2x-\sin2x)(\cos2x+\sin2x))=0$

$\Leftrightarrow (\cos x-\sin x)\left[(\cos2x-\sin2x)(1-\sqrt{3}(\cos2x+\sin2x) {} \right]=0$

$\Leftrightarrow (\cos x-\sin x)(\cos2x-\sin2x)(1-\sqrt{6}\sin(x+\frac{\pi}{4}))=0$

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