# Toán 11[Chuyên đề] Lượng giác ver.4

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#### lovelycat_handoi95

Bài 184:

$$\blue{ 1.cos10x +2cos4x^2 + 6cosx.cos3x = cosx+8cosx.cos^33x \\2.cos{\frac{4x}{3}}=cos^22x \\ 3.sin^3(x+\frac{\pi}{4})=\sqrt{2}sinx \\ 4.sin3xcos7x=sin13xcos17x$$

Bài 185:

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#### ngocthao1995

[TEX]\Leftrightarrow (sin^8x-2sin^{10}x)+(cos^8x-2cos^{10}x)=\frac{5}{4}cos2x[/TEX]

[TEX]\Leftrightarrow sin^8x(1-sin^2x)-cos^8x(-1+2cos^2x)=\frac{5}{4}cos2x[/TEX]

[TEX]\Leftrightarrow sin^8xcos2x-cos^8xcos2x=\frac{5}{4}cos2x[/TEX]

[TEX]\Leftrightarrow4 cos2x(sin^8x-cos^8x)=5cos2x[/TEX]

[TEX]\Leftrightarrow \left[\begin{cos2x=0}\\{4(sin^8x-cos^8x)=5} [/TEX]

[TEX]\Leftrightarrow \left[\begin{cos2x=0}\\{4[(sin^4x+cos^4x)(sin^4x-cos^4x)]=0} [/TEX]

[TEX]\Leftrightarrow \left[\begin{cos2x=0}\\{4(1-\frac{1}{2}sin^22x)=5(VN)} [/TEX]

[TEX]\Leftrightarrow x=\frac{\pi}{4}+k\frac{\pi}{2}[/TEX]

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#### l94

$$4.sin3xcos7x=sin13xcos17x$$

$$\Leftrightarrow sin10x-sin4x=sin30x-sin4x$$
$$\Leftrightarrow sin10x=sin30x$$
$$\Leftrightarrow .........$$
2.$$cos{4x}{3}=cos^22x$$
$$\Leftrightarrow 2cos{\frac{4x}{3}}=1+cos4x$$
$$\Leftrightarrow 2cos{\frac{4x}{3}}=1+4cos^3\frac{4x}{3}-3cos .\frac{4x}{3}$$
đến đây thì dễ r`

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#### ngocthao1995

$$3.sin^3(x+\frac{\pi}{4})=\sqrt{2}sinx$$

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

[TEX]\Leftrightarrow (sinx+cosx)^3=2sinx[/TEX]

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

Chia 2 vế của pt cho [TEX]cos^3x[/TEX] ta được

[TEX] -tan^3x+3tan^2x+tanx=0[/TEX]

.......

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#### l94

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#### miko_tinhnghich_dangyeu

[TEX] sin^8 x + cos^8 x = \frac{5cos2x}{4} + 2(sin^{10} x + cos^{10} x)[/TEX]
[TEX]\Leftrightarrow sin^8x(1-2sin^2x)+cos^8x(1-2cos^2x)-\frac{5cos2x}{4}=0[/TEX]
[TEX]\Leftrightarrow cos2x(cos^8x-sin^8x-\frac{5}{4})=0 [/TEX]

[TEX] cos^8x-sin^8x\leq cos^2x+sin^2x=1<\frac{5}{4}(VN) [/TEX]

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#### l94

Bài 184:

$$cos10x +2cos4x^2 + 6cosx.cos3x = cosx+8cosx.cos^33x$$
Bài gì mà rút gọn tuốt hết vậy nàng.
$$\large \Leftrightarrow cos10x+1+cos8x+6cosxcos3x=cosx+8cosxcos^33x$$
$$\large \Leftrightarrow 2cos9xcosx+1+6cosxcos3x=cosx+8cosxcos^33x$$
$$\large \Leftrightarrow 8cos^33xcosx-6cos3xcosx+1+6cosxcos3x=cosx+8cosxcos^33x$$
$$\large \Leftrightarrow cosx=1 \Leftrightarrow x=k2\pi$$

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#### trang_1995

Bài 186
[TEX]1.3 sin^4x+2cos^23x+cos3x=3cos^4x-cosx+1[/TEX]

[TEX]2.cos3x+6sinx=3[/TEX]

[TEX]3.3(sin^3\frac{x}{2}-cos^3\frac{x}{2})=2cosx+\frac{1}{2}sin2x[/TEX]

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#### pe_kho_12412

Một số câu trích trong các đề thi thử gần đây của trưòng mình:

1,
[TEX] 2.\sqrt{2}.cos2x + sin2x .cos (x+\frac{3{\pi}}{4})-4 sin(x+\frac{\pi}{4})=0[/TEX]

2.
[TEX] sin^2 3x - cos^2 4x = sin^2 5x - cos ^2 6x[/TEX]

3
tìm nghiệm trên khoảng [TEX]( 0; \frac{\pi}{2})[/TEX] của pt:

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

đang còn nữa, mà cũng không biết mấy câu này đã có trong pic chưa ,mình chỉ post lên các bạn thử làm thôi

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#### ngocthao1995

1.[TEX]2\sqrt{2}cos2x+sin2xcos(x+\frac{3\pi}{4})-4sinx(x+\frac{\pi}{4})[/TEX]

[TEX]\Leftrightarrow 2\sqrt{2}cos2x+sin2x(\frac{-1}{\sqrt{2}}cosx-\frac{1}{\sqrt{2}}sinx)-4(\frac{1}{\sqrt{2}}sinx+\frac{1}{\sqrt{2}}cosx)[/TEX]

[TEX]\Leftrightarrow 4cos2x-sin2xcosx-sinxsin2x-4sinx-4cosx=0[/TEX]

[TEX]\Leftrightarrow 4(cosx-sinx)(cosx+sinx)-sin2x(sinx+cosx)-4(sinx+cosx)=0[/TEX]

[TEX]\Leftrightarrow (sinx+cosx)[4(cosx-sinx)-sin2x-4]=0[/TEX]
........
[TEX]2.sin^23x-cos^24x=sin^25x-cos^26x[/TEX]

[TEX]\Leftrightarrow \frac{1-cos6x}{2}-\frac{1+cos8x}{2}=\frac{1-cos10x}{2}-\frac{1+cos12x}{2}[/TEX]

[TEX]\Leftrightarrow cos12x-cos6x+cos10x-cos8x=0[/TEX]

[TEX]\Leftrightarrow -2sin9xsin3x-2sin9xsinx=0[/TEX]

[TEX]\Leftrightarrow -2sin9x(sin3x+sinx)=0[/TEX]

....

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#### miko_tinhnghich_dangyeu

Bài 187 :
1,[TEX]4cos^4x-cos2x-\frac{1}{4}cos4x+cos{\frac{3x}{2}}=\frac{7}{2}[/TEX]
2, [TEX] 2sinx -sinx+2(cosx-1)(\sqrt[]{2}-\sqrt[]{3}cosx)=0[/TEX]
3,[TEX]sin4x(cosx-2sin4x)+cos4x(1+sinx-2cos4x)=0[/TEX]
4, [TEX]2sin^2(\frac{\pi}{2}cos^2x)=1-cos(\pi sin2x)[/TEX]
4, [TEX]cos2x+cos5x-sinx-cos8x=sin10x[/TEX]

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#### lovelycat_handoi95

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#### lovelycat_handoi95

Bài 186
[TEX]1.3 sin^4x+2cos^23x+cos3x=3cos^4x-cosx+1[/TEX]

[TEX]PT \Leftrightarrow 3(cos^4x-sin^4x)-(2cos^23x-1)-cos3x-cosx=0 \\ \Leftrightarrow 3cos2x-cos6x-2cos2xcosx=0 \\ \Leftrightarrow 3cos2x-4cos^32x+3cos2x-2cos2xcosx=0 \\ \Leftrightarrow cos2x(2cos^22x+cosx-3)=0\\ \Leftrightarrow cos2x[2(2cos^2x-1)^2+cosx-3]=0\\ [/TEX]

[TEX]2.cos3x+6sinx=3[/TEX]

[TEX]PT \Leftrightarrow cosx(4cos^2x-3)=3(1-2sinx)\\ \Leftrightarrow cosx(1-4sin^2x)=3(1-2sinx)\\ \Leftrightarrow (1-2sinx)(cosx+2sinxcosx-3)=0[/TEX]

[TEX]3.3(sin^3\frac{x}{2}-cos^3\frac{x}{2})=2cosx+\frac{1}{2}sin2x[/TEX]

[TEX]PT \Leftrightarrow 3(sin{\frac{x}{2}}-cos{\frac{x}{2}})(1+\frac{1}{2}sinx)=2cosx(1+\frac{1}{2}sinx) \\ \Leftrightarrow (1+\frac{1}{2}sinx)[3(sin{\frac{x}{2}}-cos{\frac{x}{2}})-2cosx]=0[/TEX]

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#### pe_kho_12412

tiếp nha:
Bài 188
1. [TEX]sin2x + sinx -\frac{1}{2sinx}-\frac{1}{sinx}= 2 cot2x[/TEX]

2. [TEX]\frac{3sin2x - 2 sinx}{sin2x . cosx} =2[/TEX]

3. [TEX]cos2x + 5 = 2( 2- cosx ) ( sinx - cosx)[/TEX]

4. [TEX]sinx. tan2x+ \sqrt{3}( sinx - \sqrt{3}tan2x)=3\sqrt{3}[/TEX]

5.[TEX]cos3x .cos^3 x - sin3x .sin^3x= \frac{2+3\sqrt2}{8}[/TEX]

6.[TEX] 9 sinx + 6 cosx - 3sin2x + cos2x =8[/TEX]

Bạn ghi tên bài vào nhá ^^

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#### ngocthao1995

tiếp nha:

3. [TEX]cos2x + 5 = 2( 2- cosx ) ( sinx - cosx)[/TEX]

[TEX]\Leftrightarrow (cos^2x-sin^2x)+5=2(2-cosx)(sinx-cosx)[/TEX]

[TEX]\Leftrightarrow (sinx-cosx)[2(2-cosx)+(sinx+cosx)]-5=0[/TEX]

[TEX]\Leftrightarrow (sinx-cosx)(sinx-cosx+4]-5=0[/TEX]

Đặt$$t=sinx-cosx (t\leq \sqrt{2})$$

Giải tìm t --> nghiệm

2. [TEX]\frac{3sin2x - 2 sinx}{sin2x . cosx} =2[/TEX]
Đk...

PT [TEX]\Leftrightarrow 3sin2x-2sinx=2sin2xcosx[/TEX]

[TEX]\Leftrightarrow 6sinxcosx-2sinx=4sinxcos^2x[/TEX]

[TEX]\Leftrightarrow 2sinx(3cosx-1-2cos^2x)=0[/TEX]

$$\left[\begin{sinx=0}\\{cosx = 1}\\{cosx=\frac{1}{2}}$$

....

B

#### buimaihuong

bài 189

[TEX]cotx+cos7x = 0[/TEX]

giúp hương câu này nhé

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#### trang_1995

tiếp nha:
Bài 188
5.[TEX]cos3x .cos^3 x - sin3x .sin^3x= \frac{2+3\sqrt2}{8}[/TEX]
[TEX]VT = (4cos^3x-3cox)cos^3x-(3sinx-4sin^3x)*sin^3x[/TEX][TEX]=4cos^6x-3cos^4x-3sin^4x+4sin^6x[/TEX]
[TEX]=4(cos^6x+sin^6x)-3(cos^4x+sin^4x)[/TEX]
[TEX]=4*(1-\frac{3}{4}*sin^22x)-3(1-\frac{1}{2}*sin^22x)[/TEX]
[TEX]=1-\frac{3}{2}*sin^22x[/TEX]
[TEX]=\frac{1}{4}*(3cos4x+1)[/TEX]
[TEX]\Rightarrow\frac{1}{4}(3cos4x+1)=\frac{2+3\sqrt{2}}{8}[/TEX]
[TEX]\Leftrightarrow cos4x=\frac{sqrt{2}}{2}[/TEX]

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#### lovelycat_handoi95

tiếp nha:
Bài 188

4. [TEX]sinx. tan2x+ \sqrt{3}( sinx - \sqrt{3}tan2x)=3\sqrt{3}[/TEX]

[TEX]\Leftrightarrow (sinx-3)(tan2x+\sqrt{3})=0[/TEX]

6.[TEX] 9 sinx + 6 cosx - 3sin2x + cos2x =8[/TEX]

[TEX]\Leftrightarrow 9sinx + 6cosx - 6sinxcox+ 2cos^2x-1=8[/TEX]

[TEX]\Leftrightarrow 6cosx (1-sinx)+ 2 (1-sin^2x)-9(1-sinx)=0[/TEX]

[TEX]\Leftrightarrow 6cosx (1-sinx) +2(1-sinx)(1+sinx) -9(1-sinx)=0[/TEX]

[TEX]\Leftrightarrow \left[sinx=1\\6cosx+2sinx=7(vonhiem)[/TEX]

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#### lovelycat_handoi95

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]

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