[Toán 11]Giải pt lượng giác

C

connguoivietnam

c. cos 4x/3 = cos^2 x/3
cos4x=cos^2x
2cos4x=1+cos4x
cos4x=1
e. 3(sin x + cos x) + 2sin2x -3 =0
3(sinx+cosx)+4sinxcosx-3=0
đặt sinx+cosx=t (-căn2<=t<=căn2)
vậy 2sinxcosx=t^2-1
ta có 3t+2(t^2-1)-3=0
2t^2+3t-5=0
t=1(T/M)
t=(-5/2)(loại)
sinx+cosx=1
căn2sin(x+pi/4)=1
 
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C

connguoivietnam

b. sin2x/cosx +cos2x/sinx = tan x- cotx
(sin2x/cosx)+(cos2x/sinx)=(sinx/cosx)-(cosx/sinx)
((sin2x-sinx)/cosx)+((cos2x+cosx)/sinx)=0
sinx(sin2x-sinx)+cosx(cos2x+cosx)=0
2sin^2x.cosx-sin^2x+cos^2x+cos2xcosx=0
cosx(2sin^2x+cos2x)+cos2x=0
cosx+cos2x=0
cosx+2cos^2x-1=0
 
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C

connguoivietnam

a. 9sin x +6cos x +cos 2x - 3sin2x =8
9sinx+6cosx+1-2sin^2x-6sinxcosx=8
6cosx(1-sinx)-2sin^2x+9sinx-7=0
6cosx(1-sinx)+(1-sinx)(sinx-7/2)=0
(1-sinx)(6cosx+sinx+7/2)=0
 
C

connguoivietnam

f. sin x + sin2x + sin3x = cosx + cos 2x + cos 3x
2sin2xcosx+sin2x=2cos2xcosx+cos2x
sin2x(2cosx+1)=cos2x(2cosx+1)
(2cosx+1)(sin2x-cos2x)=0
 
H

hahaha000

mấy bài này dễ quá thử nè:
tg^2 x+tg^2 y+ cotg^2 (x+y)=1
sin4x(cosx-2sin4x)+cos4x(1+sinx-2cos4x)=0
sin^3x + 4cos^3x=3cosx
sin^8 +cos^8 =32(sin^18 x+ cos^18 x)
 
B

bupbexulanxang

[TEX]tg^2 x+tg^2 y+ cotg^2 (x+y)=1[/TEX]

[TEX]sin4x(cosx-2sin4x)+cos4x(1+sinx-2cos4x)=0[/TEX]

[TEX]sin^3x + 4cos^3x=3cosx[/TEX]

[TEX]sin^8 +cos^8 =32(sin^{18} x+ cos^{18} x)[/TEX]


Cho vào TEX nhé! .
 
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B

botvit

bupbexulanxang said:
[TEX]tg^2 x+tg^2 y+ cotg^2 (x+y)=1[/TEX]

[TEX]sin4x(cosx-2sin4x)+cos4x(1+sinx-2cos4x)=0[/TEX]

[TEX]sin^3x + 4cos^3x=3cosx[/TEX]

[TEX]sin^8 +cos^8 =32(sin^{18} x+ cos^{18} x)[/TEX]


Cho vào TEX nhé! .
Bấm để xem đầy đủ nội dung ...
2.PT\Leftrightarrow[tex]sin4xcosx-2sin^24x+cos4x+cos4xsinx-2cos^24x=0[/tex]
\Leftrightarrow[tex]sin5x-2(sin^24x+cos^24x)+cos4x=0[/tex]
\Leftrightarrow[tex]sin5x-2+cos4x=0[/tex]
\Leftrightarrow[tex]sin5x+cos4x=2[/tex]
dến dây dánh giá dược ko nhở
 
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