Toán 11

K

khongphaibang

ta co :
1+cosx=2[cosx/2]^2
1+cos2x=2[cosx]^2
1+cos3x=2[cos3x/2]^2

tu do \Rightarrowpt tro thanh: 8[cosx/2.cosx.cos3x/2]^2=1/2
\Leftrightarrowcosx.cosx/2.cos3x/2=1/4(1)hoac cosx.cosx/2.cos3x/2=-1/4(2)


giai (1): cosx.cosx/2.cos3x/2=1/4
\Leftrightarrowcosx.[2cosx/2.cos3x/2]=1/2
\Leftrightarrowcosx.(cos2x.cosx)=1/2
\Leftrightarrowcos2x.2.cos^2x=1
\Leftrightarrowcos2x.cos^2x.2-1=0
\Leftrightarrowt^2-t+1=0 (t=cos2x ; -1\leqt\leq1)
\Leftrightarrowt=-1/2+(can5)/2(tm) hoac t=-1/2-(can5/2)(loai)

\Rightarrowx=+-1/2arccos(-1/2+(can5/2)



phuong trinh (2) vo nghiem


dung thi cam on cai nhe
 
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