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L

lovelycat_handoi95

ĐK: [TEX]cos3x, cos2x \not= 0[/TEX]

[TEX]PT \Leftrightarrow tan3x.tanx=(tan2x-tanx)(tan2x+tanx)= \frac{sin3x}{cos2xcosx}. \frac{sinx}{cos2xcosx}[/TEX]

[TEX]\Leftrightarrow cos3x=cosx.cos^22x \Leftrightarrow 4cos^2x-3=cos^22x \Leftrightarrow cos^22x-2cos2x+1=0[/TEX]

[TEX]\Leftrightarrow cos2x=1 [/TEX]
 
T

tuyn

[TEX]DK: sin3x \neq 0, sin2x \neq 0[/TEX]
[TEX]PT \Leftrightarrow \frac{3tanx-tan^3x}{1-3tan^2x}.tanx+tan^2x=( \frac{2tanx}{1-tan^2x})^2[/TEX]
Đặt t=tanx \Rightarrow OK
 
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