giup mih với

J

jet_nguyen

Gợi ý:
$$\sin^6x + \cos^6x=1+\sin 4x$$ Phuơng trình tương đương:
$$(\sin^2x + \cos^2x)(\sin^4x-\sin^2x\cos^2x+\cos^4x)=1+\sin 4x$$$$\Longleftrightarrow (\sin^2x+\cos^2x)-3\sin^2x\cos^2x=1+\sin 4x$$$$\Longleftrightarrow -3\sin^2x\cos^2x=\sin 4x$$$$\Longleftrightarrow \dfrac{3}{4}\sin^22x-2\sin 2x\cos 2x=0$$$$\Longleftrightarrow \sin 2x( \dfrac{3}{4}\sin 2x-2\cos 2x)=0$$ $$2\cos^3x+\sin x\cos x+ 1= 2(\sin x+ \cos x)$$ Phuơng trình tương đương:
$$2\cos x(\cos^2x-1)+\sin x\cos x+ 1- 2\sin x=0$$$$ \Longleftrightarrow -2\sin^2x\cos x+\sin x\cos x+ 1- 2\sin x=0$$$$ \Longleftrightarrow \sin x\cos x(1-2\sin x)+ 1- 2\sin x=0$$$$ \Longleftrightarrow (1-2\sin x)(\sin x\cos x+ 1)=0$$
 
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D

dhbk2013

Cosx.(Cos4x+2) + Cos2x.Cos3x = 0
<=> Cosx.Cos4x+ 2Cosx + Cos2x.Cos3x =0
<=> 1/2[Cos3x + Cos5x] +2Cosx + 1/2Cosx = 0
<=> Cos3x + 5Cosx = 0
<=> 4Cos^3x - 3Cosx + 5Cosx = 0
<=> 2Cosx(2Cos^2x + 1 ) =0
 
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