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C

connguoivietnam

[TEX]4sin^2x-2\sqrt{3}tanx+3tan^2x-4sinx+2=0[/TEX]

[TEX]4sin^2x-4sinx+1+3tan^2x-2\sqrt{3}tanx+1=0[/TEX]

[TEX](2sinx-1)^2+(\sqrt{3}tanx-1)^2=0[/TEX]

[TEX]2sinx=1[/TEX]

[TEX]\sqrt{3}tanx=1[/TEX]
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[TEX]\frac{1}{cosx} + \frac{1}{sin2x} + \frac{1}{sin4x}= 0[/TEX]

[TEX]sin4x[/TEX] khác [TEX]0[/TEX]

[TEX]\frac{1}{cosx} + \frac{1}{sin2x} + \frac{1}{2sin2xcos2x} = 0[/TEX]

[TEX]1+\frac{1}{2sinx}+\frac{1}{4cos2xsinx}=0[/TEX]

[TEX]4cos2xsinx+2cos2x+1=0[/TEX]

[TEX]4(1-2sin^2x)sinx+2(1-2sin^2x)+1=0[/TEX]

[TEX]4sinx-8sin^3x+2-4sin^2x+1=0[/TEX]

[TEX]8sin^3x+4sin^2x-4sinx+3=0[/TEX]
 
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B

bigbang195

dangquan94 said:
2.tan3x - tanx = -2sin2x
Bấm để xem đầy đủ nội dung ...

hay[TEX] \frac{\sin 2x}{\cos x \cos 3x}+2\sin 2x=0[/TEX]

[TEX]\sin 2x \bigg(\frac{1}{\cos x\cos 3x}+2\bigg)=0[/TEX]

[TEX]\frac{\sin 2x(1+2\cos x \cos 3x)}{ \cos x \cos 3x}=0[/TEX]

[TEX]\sin 2x=0[/TEX]

hoặc

[TEX] 1+2\cos x\cos3x=0 \rightarrow 1+\cos 2x+\cos 4x=1+\cos 2x+2\cos^2 2x-1 =0[/TEX]

dễ rồi
 
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G

girlbuon10594

[TEX]5(sinx+\frac{cos3x+sin3x}{1+2sin2x})=cos2x+3[/TEX]
Bấm để xem đầy đủ nội dung ...
Điều kiện: [TEX]1+2sin2x \neq 0[/TEX]
Ta có: [TEX]cos3x+sin3x=4cos^3x-3cosx+3sínx-4sin^3x[/TEX]
[TEX]= 4(cosx-sinx)(1-sinxcosx)-3(cosx-sinx)[/TEX]
[TEX]= (cosx-sinx)(1-4sinxcosx)[/TEX]
[TEX]=(cosx-sinx)(1-2sin2x)[/TEX]
Khi đó ptr có dạng:
[TEX]5(sinx+cosx-sinx)=cos2x+3[/TEX]
\Leftrightarrow [TEX]2cos^2x-5cosx+2=0[/TEX];)
 
N

nh0ctaitups

tan3x - tanx = -2sin2x ; dk: cos3x,cosx khác 0
<-> sin3x/cos3x - sinx/cosx = -2sin2x
<-> sin3x.cosx - cos3x.sinx = -2sin2x (bỏ mẫu luôn nha)
<-> sin2x = -2sin2x
<-> sin2x =0
 
0

037675582

minh muon mua may quyen sach luong giac de on thi dai hoc ma khong biet mua quyen gi ca.co ban nao biet sach gi hay thi cho minh biet cai.thank truoc nhe.
 
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