[Đề thi đại học lượng giác] từ 2002

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botvit

câu31:
PT\Leftrightarrow[tex]2sin^23x-2cos^24x=2sin^22x-2cos^26x[/tex]
\Leftrightarrow [tex]1-cos6x-1-cos8x=1-cos4x-1-cos12x[/tex]
\Leftrightarrow[tex]cos4x-cos6x=cos8x-cos12x[/tex]
\Leftrightarrow [tex]2sin5xsinx=2sin10xsin2x[/tex]
\Leftrightarrow [tex]sinx(sin5x-2sin10xcosx)=0[/tex]
 
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B

botvit

câu35
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[tex]sin^2(\frac{x}{2}-\frac{pi}{4})tan^2x-cos^2.\frac{x}{2}=0[/tex]
đk cosx#0\Leftrightarrow x#[tex]\frac{pi}{2}+kpi[/tex]
PT \Leftrightarrow [tex]2sin^2(\frac{x}{2}-\frac{pi}{4})tan^2x-2cos^2.\frac{x}{2}=0[/tex]
\Leftrightarrow [tex](1-sinx)sin^2x-cos^2x(cosx-1)=0[/tex]
\Leftrightarrow [tex] (1-sinx)(1-cos^2x)-(1-sin^2x)(cosx-1)=0[/tex]
\Leftrightarrow [tex] -(1-sinx)(cos^2x-1)-(1-sin^2x)(cosx-1)=0[/tex]
\Leftrightarrow [tex](1-sinx)(cosx-1)[-(cosx+1)-(1+sinx)]=0[/tex]
\Leftrightarrow [tex](1-sinx)(cosx-1)[-cosx-1-1-sinx)]=0[/tex]
\Leftrightarrow [tex](1-sinx)(cosx-1)(cosx+sinx+2)=0[/tex]
 
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B

botvit

câu 36
.......................
[tex]5sinx-2=3(1-sinx).tan^2x[/tex]
ddk cosx#0 \Leftrightarrow x# [tex]\frac{pi}{2}+kpi[/tex]
\Leftrightarrow [tex]5sinx-2=3(1-sinx).\frac{sin^2x}{1-sin^2x}[/tex]
\Leftrightarrow [tex]5sinx-2=\frac{3sin^2x}{1+sinx}[/tex]
\Leftrightarrow [tex] 2sin^2x+3sinx-2=0[/tex]
sau đó dễ rồi
 
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botvit

câu 29
.............................................
[tex]cos^3x+sin^3x+2sin^2x=1[/tex]
\Leftrightarrow [tex](cosx+sinx)(1-sinxcosx)=1-2sin^2x[/tex]
\Leftrightarrow ([tex]cosx+sinx)(1-sinxcosx)=cos^2x-sin^2x[/tex]
\Leftrightarrow [tex](cosx+sinx)[(cosx-sinx)-(1-sinxcosx)]=0[/tex]
\Leftrightarrow [tex]\sqrt[]{2}sin(x+\frac{pi}{4})[(cosx+sinx+sinxcosx-1)]=0[/tex]
\Leftrightarrow [tex]\sqrt[]{2}sin(x+\frac{pi}{4})=0[/tex]
hoặc [tex]cosx+sinx+sinxcosx-1[/tex]
đặt [tex]cosx+sinx=t(-\sqrt[]{2} \leq t \leq \sqrt[]{2})[/tex]
\Rightarrow [tex]\frac{t^2-1}{2}=sinxcosx[/tex]
 
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botvit

câu 28
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[tex]4sin^3x+4sin^2x+3sin2x+6cosx=0[/tex]
\Leftrightarrow [tex]4sin^3x+4sin^2x+6sinxcosx+6cosx=0[/tex]
\Leftrightarrow [tex]4sin^2x(sinx+1)+6cosx(sinx+1)=0[/tex]
\Leftrightarrow [tex](sinx+1)(4sin^2x+6cosx)=0[/tex]
\Leftrightarrow [tex](sinx+1)[4(1-cos^2x)+6cosx]=0[/tex]
\Leftrightarrow [tex](sinx+1)(-4cos^2x+6cosx+4)=0[/tex]
 
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botvit

câu6:
............................
[tex]2sin^22x+sin7x-1=sinx[/tex]
\Leftrightarrow [tex]sin7x-sinx=1-2sin^22x[/tex]
\Leftrightarrow [tex] -2cos4xsin3x=cos4x[/tex]
\Leftrightarrow [tex] cos4x(-2sin3x-1)=0[/tex]
 
B

botvit

câu14
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[tex]cosxcos2xsin3x=\frac{1}{4}sin2x[/tex]
\Leftrightarrow [tex]4cosxcos2xsin3x=sin2x[/tex]
\Leftrightarrow [tex] 2cosx(2cos2xsin3x-sinx)=0[/tex]
\Leftrightarrow [tex]2cos(sin5x+sinx-sinx)=0[/tex]
\Leftrightarrow [tex]2cosxsin5x=0[/tex]
 
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botvit

câu13
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[TEX]sin^6 x + cos^6 x = 2sin^2 ( x + \frac{\pi}{4}[/TEX])
\Leftrightarrow [tex]1-\frac{3}{4}sin^22x=1-cos(2x+\frac{\pi}{2})[/tex]
\Leftrightarrow [tex]1-\frac{3}{4}sin^22x=1+sin2x[/tex]
\Leftrightarrow [tex] 3sin^22x+4sin2x=0[/tex]
\Leftrightarrow[tex]sin2x(3sin2x+4)=0[/tex]
 
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B

botvit

câu12
......................................
[tex]cos3x.tan5x=sin7x[/tex]
đk cos5x#0 \Leftrightarrow x# [tex]\frac{pi}{2}+kpi[/tex]
PT\Leftrightarrow [tex]cos3x.\frac{sin5x}{cos5x}=sin7x[/tex]
\Leftrightarrow [tex]cos3x.sin5x=cos5x.sin7x[/tex]
\Leftrightarrow [tex]2cos3x.sin5x=2cos5x.sin7x[/tex]
\Leftrightarrow [tex]sin8x+sin2x=sin12x+sin2x[/tex]
\Leftrightarrow [tex] sin12x-sin8x=0[/tex]
\Leftrightarrow [tex]2cos10xsin2x=0[/tex]
 
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botvit

câu10
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[tex]cos^2x-2sinx+2=0[/tex]
\Leftrightarrow [tex]1-sin^2x-2sinx+2=0[/tex]
\Leftrightarrow[tex] -sin^2x-2sinx+3=0[/tex]
Bình ơi còn câu nào chưa làm thì đánh dấu = màu đỏ đi cho dễ nhín
 
H

hoangtu_faithful

botvit said:
câu31:
PT\Leftrightarrow[tex]2sin^23x-2cos^24x=2sin^22x-2cos^26x[/tex]
\Leftrightarrow [tex]1-cos6x-1-cos8x=1-cos4x-1-cos12x[/tex]
\Leftrightarrow[tex]cos4x-cos6x=cos8x-cos12x[/tex]
\Leftrightarrow [TEX]2sin5xsinx=2sin10 sin x [COLOR="Red"]Chỗ này là sin 2x[/COLOR] \Leftrightarrow [tex]sinx(sin5x-sin10x)=0[/tex]
Bấm để xem đầy đủ nội dung ...


...........................................................................
 
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S

silvery21

câu 1

[TEX] \frac{1}{sinx} + \frac{1}{sin(x-\frac{3\pi}{2})} = 4sin(\frac{7\pi}{4}-x)[/TEX]

[TEX]\frac{1}{sin(x-\frac{3\pi}{2})}=\frac{1}{cosx}[/TEX]

ta có

[TEX]4sin(\frac{7\pi}{4}-x)=-4\frac{\sqrt{2}}{2}(sinx+cosx)[/TEX]



pt\Leftrightarrow[TEX] \frac{1}{sinx} +\frac{1}{cosx}=-2\sqrt{2}(sinx+cosx)[/TEX]


\Leftrightarrow[tex]\frac{cosx+sinx}{sinxcosx}=-2\sqrt{2}(sinx+cosx)[/tex]

\Leftrightarrow[tex](sinx+cosx)(\frac{2}{sin2x}+2\sqrt{2})=0[/tex]
 
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B

bupbexulanxang

Câu 1 koi

Kâu này đầu tiên mờ hok ai làm:D
Em chém tạm vậy
ĐK : x#kpi ; x#pi/2 +kpi

Quy đồng lên.

PT\Leftrightarrow [TEX]sinx+ sin(x- 3pi/2) =4sinx.sin(x- 3pi/2).sin( 7pi/4 -x)[/TEX]
[TEX]( sin(7pi/4-x)= sin( pi- 7pi/4+x)=sin (x -3pi/4)[/TEX]

\Leftrightarrow sinx +[TEX]sin(x-3pi/2)[/TEX] =4 [TEX]sin(x-3pi/4).[/TEX]sinx.[TEX]sin(x-3pi/2)[/[/TEX]

\Leftrightarrow [TEX]sinx+sinx.cos(3pi/2) -cosx.sin(3pi/2) =4sin(x-3pi/4).sinx.[sinx.cos(3pi/2) -cosx.sin(3pi/2)][/TEX]

\Leftrightarrow [TEX]sinx +cosx =4.[sinx. cos(3pi/4) -cosx .sin(3pi/4) ] .sinx.cosx[/TEX]
\Leftrightarrow sinx +cosx = -4(sinx .[TEX]\sqrt[2]{2}/2[/TEX] +cosx .[TEX]\sqrt[2]{2}/2[/TEX] ].sinx.cosx

\Leftrightarrow sinx+cosx =-2[TEX]\sqrt[2]{2}[/TEX](sinx+cosx).sinx.cosx

\Leftrightarrow (sinx+cosx)(1-[TEX]\sqrt[2]{2}[/TEX]sin2x)=0

\Leftrightarrow [TEX]\left[\begin{sinx+cosx=0}\\{sin2x=\sqrt[2]{2}/2}[/TEX]

Tới đây giải bt nha
 
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B

bupbexulanxang

Bn viết khó hiểu y'
nhìn ko ra tưởng làm bài nào:|

Mà cách làm khó hiểu quá
Tạm chấp nhận cách tui koi:|
 
L

lanthuong12

pt : sin^6 x + cos^6 x = 2sin^2 (x+pi/4)

<=> 1- 3sin^2 x.cos^2 x = 1- cos(2x + pi/2)
<=> 3sin^2 2x + sin2x =0
<=> sin2x = 0
hoặc sin2x = -1/3

đến đây thì bạn tự giải nha!
 
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