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giải phương trình $8sinxcos3x.sin(\frac{\pi}{3}-x).sin(\frac{\pi}{3}+x)=1$

phihuyen97 · Answer 1 · 12/29/2014 9:24:53 PM

$8sinxcos3xsin(\frac{\Pi }{3}-x)sin(\frac{\Pi }{3}+x)=1$

$\Leftrightarrow 2(sin(-2x)+sin4x)(cos(-2x)-cos\frac{2\Pi }{3})=1$

$\Leftrightarrow 2(sin4x-sin2x)(cos2x+\frac{1}{2})=1$

$\Leftrightarrow 2sin4xcos2x+sin4x-2sin2xcos2x-sin2x=1$

$\Leftrightarrow sin2x+sin6x+sin4x-sin4x-sin2x=1$

$\Leftrightarrow sin6x=1$

$\Leftrightarrow 6x=\frac{\Pi }{2}+k2\Pi $

$\Leftrightarrow x=\frac{\Pi }{12}+k\frac{2\Pi }{3}$

Trả lời 29-12-14 09:24 PM

phihuyen97
1

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score 0 · Answer 2

Ta có: $2.2sin(\frac{\pi}{3}-x).sin(\frac{\pi}{3}+x)=2cos2x+1$

$2sinx.cos3x=sin4x-sin2x$

$\Rightarrow VT=(2cos2x+1)(sin4x-sin2x)=2cos2x.sin4x-sin2x=sin6x+sin2x-sin2x=sin6x$

Pt $\Leftrightarrow sin6x=1$

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