# Toán 11[Chuyên đề] Lượng giác ver.4

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#### trang_1995

Bài 190
1.[TEX]\frac{3sin2x+2cos2x+12sinx-3cosx-7}{2cosx-\sqrt{3}}=0[/TEX]
2.[TEX]2sin^2x+\sqrt{3}sin2x+1=\sqrt{3}sinx+cosx[/TEX]

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#### ngocthao1995

Một số câu LG trích trong đề thi đại học.Mọi người làm thử

Bài 191

[TEX]1.\frac{sin2x+2cosx-sinx-1}{tanx+\sqrt{3}}=0 (D-2011)[/TEX]

[TEX]2.\frac{2(cos^6x+sin^6x)-sinxcosx}{\sqrt{2}-2sinx}=0(A-2006)[/TEX]

[TEX]3.sin2xcosx+sinxcosx=cos2x+sinx+cosx(B-2011)[/TEX]

[TEX]4.sinx+cosxsin2x+\sqrt{3}cos3x=2(cos4x+sin^3x)(B-2009)[/TEX]

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#### lovelycat_handoi95

Một số câu LG trích trong đề thi đại học.Mọi người làm thử

Bài 191

[TEX]1.\frac{sin2x+2cosx-sinx-1}{tanx+\sqrt{3}}=0 (D-2011)[/TEX]

[TEX]DK : tanx \not = \sqrt{3}; cosx \not = 0 \\ PT \Leftrightarrow sin2x+2cosx-1-sinx =0 \\ \Leftrightarrow 2cosx (sinx+1) - (sinx+1)= 0 [/TEX]

[TEX]2.\frac{2(cos^6x+sin^6x)-sinxcosx}{\sqrt{2}-2sinx}=0(A-2006)[/TEX]

[TEX]DK sinx \not = \frac{ \sqrt{2}}{2} \\ PT \Leftrightarrow 2 ( cos^2x+sin^2x)(sin^4x+cos^4x-sin^2xcos^2x )-sinxcosx =0 \\ \Leftrightarrow 2[(sin^2x+cos^2x)^2-3sin^2xcos^2x]-sinxcosx=0 \\ \Leftrightarrow 6sin^2xcos^2x + sinxcosx -2 =0 \\ \Leftrightarrow \left {sinxcosx= \frac{1}{2} \\sinxcosx= -\frac{2}{3} [/TEX]

[TEX]3.sin2xcosx+sinxcosx=cos2x+sinx+cosx(B-2011)[/TEX]

[TEX]PT \Leftrightarrow 2cos^2xsinx+sinxcosx=cos2x +sinx+cosx \\ \Leftrightarrow (cos2x+1)sinx +sinxcosx = cos2x +sinx +cosx \\ \Leftrightarrow cos2x(sinx-1) + cosx(sinx-1) =0 [/TEX]

[TEX]4.sinx+cosxsin2x+\sqrt{3}cos3x=2(cos4x+sin^3x)(B-2009)[/TEX]

[TEX]\Leftrightarrow sinx+\frac{1}{2}(sin3x+sinx)+\sqrt{3}cos3x=2cos4x+ 2sin^3x \\ \Leftrightarrow \frac{1}{2}sin3x+\frac{1}{2}(3sinx-4sin^3x)+\sqrt{3}cos3x=2cos4x \\ \Leftrightarrow sin3x+\sqrt{3}cos3x=2cos4x \\ \Leftrightarrow sin(3x+\frac{\pi}{3})=cos4x[/TEX]

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#### lovelycat_handoi95

Bài 190
[TEX]2.\frac{3sin2x+2cos2x+12sinx-3cosx-7}{2cosx-\sqrt{3}}=0[/TEX]

[TEX]DK cosx \not= \sqrt{3}{2} \\ PT \Leftrightarrow 3sin2x+2cos2x+12sinx-3cosx-7 = 0 \\ \Leftrightarrow 6sinxcosx+2(1-2sin^2x)+12sinx-3cosx-7=0 \\ \Leftrightarrow 3cosx(2sinx-1) -(4sin^2x -12sinx +5 )=0 \\ \Leftrightarrow 3cosx(2sinx-1) -(2sinx-5)(2sinx-1 ) =0 \\ [/TEX]

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#### lovelycat_handoi95

Bài 190
[TEX]2.\frac{3sin2x+2cos2x+12sinx-3cosx-7}{2cosx-\sqrt{3}}=0[/TEX]

[TEX]DK cosx \not= \sqrt{3}{2} \\ PT \Leftrightarrow 3sin2x+2cos2x+12sinx-3cosx-7 = 0 \\ \Leftrightarrow 6sinxcosx+2(1-2sin^2x)+12sinx-3cosx-7=0 \\ \Leftrightarrow 3cosx(2sinx-1) -(4sin^2x -12sinx +5 )=0 \\ \Leftrightarrow 3cosx(2sinx-1) -(2sinx-5)(2sinx-1 ) =0 \\ [/TEX]

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#### nuhoangbongdem95

Bài 189 :

[TEX]1.tan2x-tan3x-tan5x=tan2xtan3xtan5x \\ 2.tan^22x-tan^23x+tan5x=tan^22xtan^23xtan5x\\ 3.cosxcos2xcos3x+sinxsin2xsin3x=1\\4.sin^4x+cos^4x={\frac{3+\sqrt{3}}{2}}sin2xcos2x+{\frac{2-3\sqrt{3}}{2}}cos^3x \\ 5.{\frac{\sqrt{2}(sinx-cosx)^2(1+2sin2x)}{sin3x+sin5x}}=1-tanx[/TEX]

1.
ĐK :
$\left\{\begin{array}{l}cos2x\neq 0 \\ cos3x\neq 0 \\ cos5x\neq 0\end{array}\right$

$\leftrightarrow \left\{\begin{array}{l}x\neq \frac{\pi}{4} +\frac{k\pi}{2} \\ x\neq \frac{\pi}{6} +\frac{k\pi}{3}\\ x\neq \frac{\pi}{10}+\frac{k\pi}{5} \end{array}\right$
$(k \in Z)$

$tan2x - tan3x - tan5x = tan2x.tan3x.tan5x$

$\leftrightarrow tan2x.tan3x.tan5x + tan5x - tan2x + tan3x= 0$

$\leftrightarrow \frac{sin2x.sin3x.sin5x}{cos2x.cos3x.cos5x} + \frac{sin3x}{cos5x.cos2x} + \frac{sin3x}{cos3x} = 0$

$\leftrightarrow \frac{sin2x.sin3x.sin5x + sin3x.cos3x + sin3x.cos5x.cos2x}{cos2x.cos3x.cos5x} = 0$

$\leftrightarrow sin3x.(sin2x.sin5x + cos3x + cos2x.cos5x) = 0$

$1. sin3x = 0 \leftrightarrow x= \frac{k\pi}{3}$
$(k \in Z)$

$2. sin2x.sin5x + cos3x + cos2x.cos5x = 0$

$\leftrightarrow \frac{1}{2}.(cos3x - cos7x) + cos3x + \frac{1}{2}.(cos3x + cos7x) = 0$

$\leftrightarrow cos3x = 0$
(loại )

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#### lovelycat_handoi95

Bài 189 :

[TEX] \\ 2.tan^22x-tan^23x+tan5x=tan^22xtan^23xtan5x\\ 3.cosxcos2xcos3x+sinxsin2xsin3x=1\\4.sin^4x+cos^4x={\frac{3+\sqrt{3}}{2}}sin2xcos2x+{\frac{2-3\sqrt{3}}{2}}cos^3x \\ 5.{\frac{\sqrt{2}(sinx-cosx)^2(1+2sin2x)}{sin3x+sin5x}}=1-tanx[/TEX]

Còn 4 bài này nữa .( .

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#### anhsao3200

Bài 2

[TEX]cosxcos2xcos3x+sinxsin2xsin3x=1\\[/TEX]

[TEX]\Leftrightarrow (cosxcos2xcos3x)^2+2(sinxcosxsin2xcos2xsin3xcos3x)+(sinxsin2xsin3x)^2=1 [/TEX]

Nhận xét

[TEX] -1 \leq cosxcos2xcos3x \leq 1[/TEX]

[TEX] -1 \leq sinxsin2xsin3x \leq1 [/TEX]

Đặt
[TEX]cosxcos2xcos3x = cos \alpha[/TEX]

[TEX]\\sinxsin2xsin3x= sin \alpha[/TEX]

[TEX]pt \Leftrightarrow sin^2\alpha +cos^2\alpha+2sin2\alpha=1[/TEX]

[TEX] Xong [/TEX]

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#### ngocthao1995

Còn 4 bài này nữa .( .

Bài 5

Đk...

PT [TEX]\Leftrightarrow \frac{\sqrt{2}(sinx-cosx)^2(1+4sinxcosx)}{2sin4xcosx}=\frac{cosx-sinx}{cosx}[/TEX]

[TEX]\Leftrightarrow \sqrt{2}cosx(sinx-cosx)^2(1+4sinxcosx)=-2sin4xcosx(sinx-cosx))[/TEX]

[TEX]\Leftrightarrow \sqrt{2}(sinx-cosx)(1+4sinxcosx)+2sin4x=0[/TEX]

[TEX]\Leftrightarrow \sqrt{2}(sinx-cosx)(1+4sinxcosx)-4sin2x(sinx-cosx)(sinx+cosx)=0[/TEX]

[TEX]\Leftrightarrow (sinx-cosx)(\sqrt{2}+4\sqrt{2}sinxcosx-4sin2x(sinx+cosx)=0[/TEX]

[TEX]\left[\begin{sinx-cosx=0}\\{\sqrt{2}+4\sqrt{2}sinxcosx-4sinxcosx(sinx+cosx)=0(2)} [/TEX]

[TEX](2)[/TEX] Đặt [TEX]sinx+cosx=t \Leftrightarrow sinxcosx=\frac{t^2-1}{2}[/TEX]
.....

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#### li94

[TEX]2sin^2x - sin2x + sinx + cosx - 1 = 0[/TEX] .

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#### ngocthao1995

[TEX]2sin^2x - sin2x + sinx + cosx - 1 = 0[/TEX] .

[TEX]\Leftrightarrow-(1-2sin^2x)-sin2x+sinx+cosx=0[/TEX]

[TEX]\Leftrightarrow -cos2x-sin2x+sinx+cosx=0[/TEX]

[TEX]\Leftrightarrow cos2x+sin2x=cosx+sinx[/TEX]

[TEX]\Leftrightarrow cos(\frac{\pi}{4}-2x)=cos(\frac{\pi}{4}-x)[/TEX]

........

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#### miko_tinhnghich_dangyeu

Bài 193

1,[TEX]4cos^4x-cos2x-\frac{1}{4}cos4x+cos{\frac{3x}{2}}=\frac{7}{2}[/TEX]

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

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#### lovelycat_nd9x

Bài 189 :

[TEX]\\ 2.tan^22x-tan^23x+tan5x=tan^22xtan^23xtan5x[/TEX]

[TEX]DK : \left{cos2x \not=0 \\ cos3x\not=0 \\ cos5x\not =0 [/TEX]

[TEX]PT 2 \Leftrightarrow tan5x(1-tan^22xtan^23x)=tan^23x-tan^22x \\ \Leftrightarrow tan5x(\frac{cos^22xcos^23x-sin^22xsin^23x}{cos^23xcos^22x})=(tan3x-tan2x)(tan3x+tan2x) \\ \Leftrightarrow tan5x(\frac{cos^22xcos^23x-sin^22xsin^23x}{cos^23xcos^22x})= \frac {sinxsin5x}{cos^22xcos^23x}\\ \Leftrightarrow \left[sin5x=0 \\ cos^22xcos^23x-sin^22xsin^23x = sinxcos5x (2) [/TEX]

[TEX](2) \Leftrightarrow (cos2xcos3x-sin2xsin3x)(cos2xcos3x+sin2xsin3x)=sinxcos5x \\ \Leftrightarrow cosx cos5x=sinxcos5x \\ \Leftrightarrow \left[cos5x=0\\cosx=sinx[/TEX]

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#### lovelycat_nd9x

Bài 193

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

[TEX]PT \Leftrightarrow sin4xcosx-2sin^24x+cos4x+sinxcos4x-2cos^24x=0 \\ \Leftrightarrow (sin4xcosx+sinxcos4x)+cos4x -2(sin^24x+cos^24x)=0 \\ \Leftrightarrow sin5x+cos4x=2 \\ \Leftrightarrow \left{sin5x=1 \\ cos4x=1[/TEX]

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#### matnatinhyeu_1995

Bài 194:

[TEX]1) sin2x + sinx - \frac{1}{2sinx} - \frac{1}{sin2x} = 2cot2x[/TEX]

[TEX]2) 2cosx - cos^2 ( x + \pi ) = \frac{8}{3} + sin2x + 3cos(x + \frac{\pi}{2} ) +\frac{1}{3}sin^2x[/TEX]

p/s: pic giờ trầm quá :-SS

bao nhiêu cao thủ hùi trước đâu hết rồi????????:-SS

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#### lovelycat_handoi95

[TEX]1) sin2x + sinx - \frac{1}{2sinx} - \frac{1}{sin2x} = 2cot2x[/TEX]

Đề này chứ nhở

[TEX]DK: sin2x \not=0 ;sinx\not=0 \\PT \Leftrightarrow [sin2x-\frac{1}{sin2x}]+[sinx-\frac{1}{2sinx}]=2cot2x \\ \Leftrightarrow \frac{-cos^22x}{sin2x}-\frac{cos2x}{2sinx}=2.\frac{cos2x}{sin2x} \\ \Leftrightarrow \frac{cos2x}{2sinx}.(\frac{cos2x}{cosx}+1+\frac{2} {cosx})=0 \\ \Leftrightarrow \frac{cos2x}{2sinx}.(\frac{2cos^2x+cosx+1}{cosx})=0[/TEX]

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#### kysybongma

Bài 194:

[TEX]2) 2cosx - cos^2 ( x + \pi ) = \frac{8}{3} + sin2x + 3cos(x + \frac{\pi}{2} ) +\frac{1}{3}sin^2x[/TEX]

p/s: pic giờ trầm quá :-SS

bao nhiêu cao thủ hùi trước đâu hết rồi????????:-SS

\Leftrightarrow[TEX]6cosx - 3(cosx)^2 = 8 + 3sin2x - 9 sinx +(sinx)^2[/TEX]

\Leftrightarrow[TEX]6cosx - 6sinxcosx = -2(sinx)^2 - 9sinx +11 [/TEX]

\Leftrightarrow[TEX]6cosx(1-sinx) = (1-sinx)(2sinx+11)[/TEX]

\Leftrightarrow[TEX](1-cosx)(6cosx-2sinx-11)=0[/TEX]

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#### lovelycat_handoi95

Bài 195 :
(tiếp cả nhà ^^)

$$\blue{1.\frac{(1-cosx)^2+(1+cosx)^2}{4(1-sinx)}-tan^2xsinx=\frac{1+sinx}{2}+tan^2x (DHKT-96) \\ 2.\frac{sinxcot5x}{cos9x}=1 ( DH Hue -A - 99) \\ 3. tanx-sin2x-cos2x+2(2cosx-\frac{1}{cosx})=0 (DHL-98) \\ 4.\frac{sin^4x+cos^4x}{sin2x}=\frac{1}{2}(tanx+cotx)(DHBKHN-2000) \\ 5. 6sinx-2cos^3x=\frac{5sin4xcosx}{2cos2x}(DHYHN-95)$$

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#### nangbanmai360

4, ĐK x khác kpi/2
Ta thấy sin^4 x + cos^4 x = (sin^2 x+cos^2 x) - 2sin^2 xcos^2 x
tanx + cotx = (sin^2 x+cos^2 x)/ (2sinxcosx)
\Rightarrow sin2x= 0
\Rightarrow Phương trình vô nghiệm

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#### nangbanmai360

5, Đk x# pi/4 +kpi/2
\Leftrightarrow 6sinx - 2 cos^3x = 5sin2xcosx
\Leftrightarrow 3sinx - cos^3 x - 5sinxcos^2 x=0
Nếu cosx =0 là nghiệm\Rightarrow vô lí
cosx# 0 .Chia cả hai vế cho cos^3 x
\Rightarrow 3tan^3 x- 2tanx -1 =0
\Rightarrow x= pi/4 +npi
Vậy phương trình vô nghiệm

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#### heartrock_159

Bài 195 :
(tiếp cả nhà ^^)

$$\blue{1.\frac{(1-cosx)^2+(1+cosx)^2}{4(1-sinx)}-tan^2xsinx=\frac{1+sinx}{2}+tan^2x (DHKT-96) \\ 2.\frac{sinxcot5x}{cos9x}=1 ( DH Hue -A - 99) \\ 3. tanx-sin2x-cos2x+2(2cosx-\frac{1}{cosx})=0 (DHL-98) \\ 4.\frac{sin^4x+cos^4x}{sin2x}=\frac{1}{2}(tanx+cotx)(DHBKHN-2000) \\ 5. 6sinx-2cos^3x=\frac{5sin4xcosx}{2cos2x}(DHYHN-95)$$

1.Đk : [TEX]cosx <> 0[/TEX]

[TEX]\Leftrightarrow \frac{1+cos^2 x}{1 - sinx} = (sinx + 1)(2tan^2 x + 1)[/TEX]

[TEX]\Leftrightarrow 1 + cos^2x = cos^2x(2tan^2x + 1)[/TEX]

[TEX]\Leftrightarrow 1 = 2sin^2x[/TEX]

[TEX]\Leftrightarrow \[sinx = \frac{\sqrt{2}}{2} \\ sinx = - \frac{\sqrt{2}}{2}[/TEX]

2. Đk:...

[TEX]PT \Leftrightarrow sinx.cos5x=sin5x.cos9x[/TEX]

[TEX]\Leftrightarrow sin6x + sin(-4x) = sin14x + sin(-4x)[/TEX]

[TEX]\Leftrightarrow sin6x = sin14x [/TEX]

3.

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