Giúp mình với ! THanks !

G

girlbuon10594

Nếu không nhầm thì đề bài cậu muốn hỏi là:
[TEX]\frac{6cos^32x+2sin^32x}{3cos2x-sin2x}=cos4x[/TEX]

\Leftrightarrow [TEX]6cos^32x+2sin^32x=(2cos^22x-1)(3cos2x-sin2x)[/TEX]

\Leftrightarrow [TEX]6cos^32x+2sin^32x=6cos^32x-2cos^22xsin2x-3cos2x+sin2x[/TEX]

\Leftrightarrow [TEX]2sin2x(sin^22x+cos^22x)+3cos2x-sin2x=0[/TEX]


\Leftrightarrow [TEX]sin2x+3cos2x=0[/TEX];)
 
M

minh_hhc

\frac{6cos^32x+2sin^32x}{3cos2x-sin2x}=cos4x

6cos^32x+2sin^32x=(2cos^22x-1)(3cos2x-sin2x)

6cos^32x+2sin^32x=6cos^32x-2cos^22xsin2x-3cos2x+sin2x

2sin2x(sin^22x+cos^22x)+3cos2x-sin2x=0


sin2x+3cos2x=0
 
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