giai gium em bai toan nay nhe em dang rat can , thanks anh chi truoc nhe

A

anthonizan

trả lời

nếu là 4cosx - 2cos2x - cos4x = 1 thì giải là :
khi không có hướng làm tốt nhất là cứ phân tích ra:
\Leftrightarrow 4cosx - 4cosx^2 - 8cosx^4 + 8cosx^2 = 0
\Leftrightarrow(cosx - 1) (cosx + 2cosx^2 + 2cosx^3)= 0
hướng là vậy nếu có sai sót tính toán gì bỏ qua ghen!nếu đúng là cosx nhớ thank.
 
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D

daothehoanght

media.tuoitre.com.vn/download/tailieu/tuyensinh/2007/Luonggiac-Chuong8.pdf
copy rui pase vao IE coi
 
N

ngoxuanquy

gif.latex
 
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L

l94

ngoc_bn said:
[TEX]4cosx-2cos2x-cos4x=1[/TEX]
anh chi giup em nhanh len nhe cang som cang tot ,thanks nhieu
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[TEX]4cosx-(4cos^2x-2)-(2cos^2{2x}-1)=1[/TEX]

[TEX]4cosx-4cos^2x+2-[2.(cos2x)^2-1]=1[/TEX]

[TEX]4cosx-4cos^2x+2-[2.(2cos^2x-1)^2-1]=1[/TEX]

[TEX]4cosx-4cos^2x+2-[2.(4cos^4x-4cos^2x+1)-1]=1[/TEX]

[TEX]4cosx-4cos^2x+2-(8cos^4x-8cos^2x+2-1)=1[/TEX]

[TEX]4cosx-4cos^2x+2-8cos^4x+8cos^2x-2=0[/TEX]

[TEX]2cosx-2cos^2x-4cos^4x+4cos^2x=0[/TEX]

[TEX]{-2cos^4x+cos^2x+cosx+1=0}[/TEX]

[TEX]cosx(-2cos^3x+cosx+1)=0[/TEX]
 
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B

buixuanha_94

4cosx - 2cos2x - cos4x = 1 <=> 4cosx - 2cos2x - cos4x = [tex]\ sin2x^2 + {cos2x^2}[/tex]
<=> 4cosx - 2cos2x - [tex]\ cos2x^2 + {sin2x^2}[/tex] = [tex]\ sin2x^2 + {cos2x^2}[/tex] <=> 4cosx - 2cos2x - [tex]\ 2cos2x^2 [/tex] = 0 <=> 4cosx - 2cos2x( 1 + cox2x ) = 0 <=> 4cosx - 2cos2x( 1+ [tex]\ 2cosx^2 [/tex]- 1 ) = 0 <=> 4cosx - 2cos2x[tex]\2cosx^2 [/tex] = 0 <=> 4cosx( 1 - cosxcos2x ) = 0
cosx = 0 _______<=> x = pi/2 + k2pi_________ <=> x = pi/2 + k2pi
cosxcos2x= 1 ______ [tex]\ 2cosx^3 [/tex]- cosx = 1__________ x = k2pi
:)&gt;-
 
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N

nguyentankha1994py

N

namtj3nhaj_tb

buixuanha_94 said:
4cosx - 2cos2x - cos4x = 1 <=> 4cosx - 2cos2x - cos4x = [tex]\ sin2x^2 + {cos2x^2}[/tex]
<=> 4cosx - 2cos2x - [tex]\ cos2x^2 + {sin2x^2}[/tex] = [tex]\ sin2x^2 + {cos2x^2}[/tex] <=> 4cosx - 2cos2x - [tex]\ 2cos2x^2 [/tex] = 0 <=> 4cosx - 2cos2x( 1 + cox2x ) = 0 <=> 4cosx - 2cos2x( 1+ [tex]\ 2cosx^2 [/tex]- 1 ) = 0 <=> 4cosx - 2cos2x[tex]\2cosx^2 [/tex] = 0 <=> 4cosx( 1 - cosxcos2x ) = 0
cosx = 0 _______<=> x = pi/2 + k2pi_________ <=> x = pi/2 + k2pi
cosxcos2x= 1 ______ [tex]\ 2cosx^3 [/tex]- cosx = 1__________ x = k2pi
:)>-
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bj3t la` y' tg? ban dung nhung phaj la` th3' nay` chu' nhj?
4cosx - 2cos2x -([TEX]cos^2(2x)- sin^2(2x)[/TEX])=[TEX]sin^2(2x)+cos^2(2x)[/TEX].......................???:confused:
 
N

namtj3nhaj_tb

buixuanha_94 said:
4cosx - 2cos2x - cos4x = 1 <=> 4cosx - 2cos2x - cos4x = [tex]\ sin2x^2 + {cos2x^2}[/tex]
<=> 4cosx - 2cos2x - [tex]\ cos2x^2 + {sin2x^2}[/tex] = [tex]\ sin2x^2 + {cos2x^2}[/tex] <=> 4cosx - 2cos2x - [tex]\ 2cos2x^2 [/tex] = 0 <=> 4cosx - 2cos2x( 1 + cox2x ) = 0 <=> 4cosx - 2cos2x( 1+ [tex]\ 2cosx^2 [/tex]- 1 ) = 0 <=> 4cosx - 2cos2x[tex]\2cosx^2 [/tex] = 0 <=> 4cosx( 1 - cosxcos2x ) = 0
cosx = 0 _______<=> x = pi/2 + k2pi_________ <=> x = pi/2 + k2pi
cosxcos2x= 1 ______ [tex]\ 2cosx^3 [/tex]- cosx = 1__________ x = k2pi
:)>-[/Q
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