tat ca? lien wan toi luong giac...............???????????

N

ngochan92

bai2:sinx+sin2x+sin3x+sin4x+sin5x+sin6x=0
truong dh su pham vinh,1997
bai3:[tex]sin^3xcos3x+cos^3xsin3x=sin^34x[/tex]
 
B

boy_depzai_92

[tex]sinx+sin2x+sin3x+sin4x+sin5x+sin6x=0[/tex]
Bấm để xem đầy đủ nội dung ...
[tex]=sin2x+(sinx+sin3x)+sin5x+(sin4x+sin6x)=0[/tex]
[tex]=sin2x+2sin2x.cosx+sin5x+2sin5x.cosx=0[/tex]
[tex]=(sin2x+sin5x)(2cosx-1)=0[/tex]
[tex]\left[\begin{cosx=\frac{1}{2}}\\{sin2x=-sin5x}[/tex]
[tex]\left[\begin{x= \pm \frac{\pi}{6}+k.2\pi}\\{2x=5x+\pi+k.2\pi}\\{2x=5x+k.2\pi}[/tex]
 
A

apple_red

bai3:[tex]sin^3xcos3x+cos^3xsin3x=sin^34x[/tex][/QUOTE]
[TEX]pt<=> sin^3x(4cos^3x-3cosx) + cos^3x(3sinx-4sin^3x)=sin^34x[/TEX]
[TEX]<=> 3sinxcosx(cos^2x-sin^2x)=sin^34x[/TEX]
[TEX]<=> \frac{3}{2}sin2xcos2x=sin^34x[/TEX]
[TEX]<=> \frac{3}{4}sin4x=sin^34x[/TEX]
[TEX]<=> 3sin4x-4sin^34x=0[/TEX]
[TEX]<=> sin12x=0[/TEX]
 
L

lan_anh_a

ngochan92 said:
bai3:[tex]sin^3xcos3x+cos^3xsin3x=sin^34x[/tex]
Bấm để xem đầy đủ nội dung ...


VT = [TEX]\frac{1}{4} ( 3 sinx - sin 3x) cos3x + \frac{1}{4} (3 cosx + cos3x) sin 3x[/TEX]


[TEX]= \frac{3}{4} ( sin x.cos 3x + cosx.sin 3x) = \frac{3}{4} sin 4x[/TEX]

==> [TEX]pt <=> 3 sin 4x - 4 sin ^3 4x = 0[/TEX]

[TEX]\Leftrightarrow sin 12 x = 0 [/TEX]

[TEX]\Leftrightarrow 12x = k\pi \Leftrightarrow x = \frac{k\pi}{12} ( k\in Z)[/TEX]
 
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