Lượng giác

[viva_la_vida]

[TEX]8cos^3(x+\frac{\pi}{3})=cos3x[/TEX]

Đặt [TEX]x+\frac{\pi}{3}=t \Rightarrow 3x=3t-\pi[/TEX]

pt trở thành [TEX]8cos^3t=cos(3t- \pi )[/TEX]

[TEX]\Leftrightarrow 8cos^3t+cos3t=0[/TEX]

[TEX]\Leftrightarrow 8cos^3t+4cos^3t-3cost=0[/TEX]

[TEX]\Leftrightarrow cost(4cos^2t-1)=0[/TEX]

 

[kiburkid]

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[TEX]cos^2(x+\frac{\pi}{3}) + cos^2(x+\frac{2\pi}{3}) = \frac{sinx+1}{2}[/TEX]

 

[lantrinh93]

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[TEX]cos^2(x+\frac{\pi}{3}) + cos^2(x+\frac{2\pi}{3}) = \frac{sinx+1}{2}[/TEX]

bài này sao chị làm lẽ quá nhĩ
\Leftrightarrow
[TEX]2+ cos(2x+\frac{2.\pi }{3})+cos(2x+4.\frac{\pi }{3})= sinx+1[/TEX]


<..> [TEX] -cos2x=sinx+1 [/TEX]
[TEX] -1+2(sinx)^2=sinx+1 [/TEX]
.>[TEX] 2(sinx)^2-sinx-2=0[/TEX]
(
nhầm chổ nào nhĩ

 

[mr.duchop]

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[TEX]cos^2(x+\frac{\pi}{3}) + cos^2(x+\frac{2\pi}{3}) = \frac{sinx+1}{2}[/TEX]

[TEX](cosxcos\frac{\pi }{3} - sinxsin\frac{\pi }{3})^2 +(cosxcos\frac{2\pi }{3} - sinxsin\frac{2\pi }{3})^2 = \frac{sinx+1}{2}[/TEX]

[TEX]\Rightarrow[/TEX] [TEX]\frac{1}{4}(cos^2x- 2\sqrt{3}sinx + 3in^2x)+\frac{1}{4}(cos^2x+ 2\sqrt{3}sinx + 3in^2x)= \frac{sinx+1}{2}[/TEX]

[TEX]\Leftrightarrow[/TEX][TEX] cos^2x + 3sin^2x = sinx + 1[/TEX]

[TEX]\Leftrightarrow[/TEX][TEX] 2sin^2x + sinx = 0[/TEX]

[TEX]\Rightarrow[/TEX] ............

 

[mr.duchop]

[TEX]sin^4x + sin^4(x + \frac{\pi }{4}) + sin^4(x - \frac{\pi }{4})=\frac{9}{8}[/TEX]

 

[heongoc_97]

[TEX]sin^4x + sin^4(x + \frac{\pi }{4}) + sin^4(x - \frac{\pi }{4})=\frac{9}{8}[/TEX]

[TEX]<=>8sin^4x+2(1+sin2x)^2+2(1-sin2x)^2=9[/TEX]
[TEX]<=>8sin^4x+2+4sin2x+sin^22x+2-4sin2x+sin^22x=9[/TEX]
[TEX]<=>8sin^4x+2sin^22x=5[/TEX]
[TEX]<=>8sin^4x+2(1-cos^22x)=5[/TEX]
[TEX]<=>8sin^4x-2cos^22x=3[/TEX]
[TEX]<=>8sin^4x-2(1-2sin^2x)^2=3[/TEX]
[TEX]<=>8sin^4x-2+8sin^2x-8sin^4x=3[/TEX]
[TEX]<=>8sin^2x-5=0[/TEX]

 

[mr.duchop]

dòng 3->4 :|
_____________________________________________________________

 

[kiburkid]

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[giotsuong_93]

[TEX]\sqrt3(-sin^2x+cosx)+3sinx-sinxcos=0 \Leftrightarrow-sinx(sqrt3sinx+cosx)+sqrt3(cosx+sqrt3sinx)=0 \Leftrightarrow(sqrt3sinx+cosx)(sqrt3-sinx)=0[/TEX]
tớ rất cảm ơn+ ủng hộ topic

sai [TEX]\sqrt3(-sin^2x+cosx)+3sinx-sinxcos=0[/TEX]

[TEX]\sqrt3(-2sin^2x+cosx)+3sinx-sinxcos=0[/TEX]

 

[kiburkid]

[TEX]sin3x + cos3x - 2\sqrt2cos(x+\frac{\pi}{4}) + 1 = 0[/TEX]

 

[lamtrang0708]

pt trở thành
[tex]3sinx-4sin^3x +4cos^3x-3cosx-2(cosx-sinx)+1=0 [/tex]
[tex]\Leftrightarrow 3(sinx-cosx)-4(sinx-cosx)(1+sinxcosx)+2(sinx-cosx)+1=0 [/tex]
đặt sinx-cosx=t
thì [tex]sinxcosx= (1-t^2)/2 [/tex]
ra pt đôí xứng

 

[kiburkid]

Em làm con ni cả buổi chiều ko ra
Nản ghê :|


[TEX]2cos3x (2cos2x+1) = 1[/TEX]