[toán 11]Giải phương trình lg

L

lovelycat_handoi95

[TEX] DK : sin2x \not = 0 \\ PT \Leftrightarrow \frac{cos2x}{sinxcosx}+4sin2x=\frac{2}{sin2x} \\ \Leftrightarrow 2cos2x+4sin^22x=2 \\ \Leftrightarrow cos2x+2(1-cos^22x)=1 \\ \Leftrightarrow 2cos^22x-cos2x-1 =0 \\ \Leftrightarrow \left[cos2x=1\\cos2x= \frac{-1}{2}[/TEX]
 
K

khunjck

[TEX]cotx-tanx+4.sin2x=\frac{2}{sin2x}[/TEX] (1)

Đk: [TEX]sin2x\not= \0[/TEX]

\Leftrightarrow[TEX]\frac{cosx}{sinx}-\frac{sinx}{cosx}+4.sin2x=\frac{2}{2.sinx.cosx}[/TEX]
\Leftrightarrow[TEX]cos^2x-sin^2x+8.sin^2x.cos^2x=1[/TEX]
\Leftrightarrow[TEX]cos^2x-(1-cos^2x)+8.(1-cos^2x).cos^2x-1=0[/TEX]
\Leftrightarrow[TEX]4.cos^4x-5.cos^2x+1=0[/TEX]
 
You must log in or register to reply here.
Top Bottom