Toán 11 Phương trình lượng giác.

[lamanhnt]

4,
$$4cosx(cos^2x-2)=-3sqrt{2}sin2x$$
$$2cosx(1+sin^2x)=3sqrt{2}sinxcosx$$
$$cosx(2+2sin^2x-3sqrt{2}sinx)=0$$

 

[silvery21]

[TEX]37/ cos^4x + sin^4x + cos(x - \frac{\pi}{4})sin(3x - \frac{\pi}{4}) - \frac{3}{2} = 0[/TEX]

[TEX]39/ 4cos^3x + 3\sqrt{2}sin2x = 8cosx[/TEX]

nốt 2 câu

39

pt \Leftrightarrow [TEX]2 cos x( 2 cos^2 x +3\sqrt{2} sin x -4)=0[/TEX]

[TEX]done!!!!!!!!![/TEX]

37: hạ bậc ; áp dụng [TEX]cos. sin =.........[/TEX]

pt\Leftrightarrow [TEX]2 + 2 cos^2 2x + 2 sin 2x - 2 cos4x -6=0[/TEX]

\Leftrightarrow [TEX]2 sin ^2 2x + 2 sin 2x -4 =0[/TEX]

[TEX] done[/TEX]...........................

 

[rua_it]

[TEX]37/ cos^4x + sin^4x + cos(x - \frac{\pi}{4})sin(3x - \frac{\pi}{4}) - \frac{3}{2} = 0[/TEX]
Các bạn vào ủng hộ nha!

[tex]cos^4x + sin^4x + cos(x - \frac{\pi}{4})sin(3x - \frac{\pi}{4}) - \frac{3}{2} = 0[/TEX]

$$\Rightarrow (sin^2x+cos^2x)^2-2sin^2x.cos^2x+\frac{1}{2}.(sin2x-cos4x)-\frac{3}{2}=0$$

$$\Rightarrow -\frac{1}{2}-\frac{1}{2}.sin^22x+\frac{1}{2}.(sin2x-cos4x)=0$$

$$\Rightarrow -\frac{1}{2}-\frac{1}{2}.sin^22x-\frac{1}{2}.(1-2sin^22x)=0$$

$$\Rightarrow-1-sin^22x-1-2sin^22x=0$$

$$\Rightarrow sin^22x-sin2x-2=0$$

Đi học đã

 

[rua_it]

[TEX]36/ cos^23xcos2x - cos^2x = 0[/TEX]
Các bạn vào ủng hộ nha!

[tex]\huge cos^23xcos2x - cos^2x = 0[/TEX]

$$\huge \Rightarrow \frac{cos2x.(cos6x+1)}{2}-\frac{cos2x+1}{2}=0$$

$$\huge \Rightarrow cos6x.cos2x=1$$

$$\huge \Rightarrow cos8x+cos4x=2$$

[tex]\huge \Rightarrow \left{\begin{cos^24x=1}\\{cos4x=1}[/tex]

 

[rua_it]

[TEX]40/ cos(2x + \frac{\pi}{4}) + cos(2x - \frac{\pi}{4}) + 4sinx = 2 + \sqrt{2}(1 - sinx)[/TEX]

Các bạn vào ủng hộ nha!

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

$$\Rightarrow 2.cos2x.cos.\frac{\pi}{4}+4sinx-2-\sqrt{2}.(1-sinx)=0$$

$$\Rightarrow (1-2sin^2x).\sqrt{2}+(2.\sqrt{2}+1).\sqrt{2}.sinx-\sqrt{2}.(\sqrt{2}+1)=0$$

$$\Rightarrow -2.\sqrt{2}.sin^2x+(2.\sqrt{2}+1).\sqrt{2}.sinx-2=0$$

$$\Rightarrow 2.sin^2x-(2\sqrt{2}+1)sinx+\sqrt{2}=0$$

 

[djbirurn9x]

Tiếp tục nào

[TEX]41/ cotx - 1 = \frac{cos2x}{1 + tanx} + sin^2x - \frac{1}{2}sin2x[/TEX]

[TEX]42/ cotx - tanx + 4sin2x = \frac{2}{sin2x}[/TEX]

[TEX]43/ \frac{3sin3x - \sqrt{3}cos9x - 1}{sin2x + sinx} = 4sin^23x(2cosx - 1)[/TEX]

[TEX]44/ tanx - sin2x - cos2x + 2(2cosx - \frac{1}{cosx}) = 0[/TEX]

[TEX]45/ (1 + 2cos3x)sinx + sin2x = 2sin^2(2x + \frac{\pi}{4})[/TEX]

 

[silvery21]

[TEX]42/ cotx - tanx + 4sin2x = \frac{2}{sin2x}[/TEX]

đk........

[TEX]sin2x(\frac{cos^2x-sin^2x}{sinx.cosx})+4sin^22x=2[/TEX]

[TEX]\Leftrightarrow 2cos2x+4(1-cos^22x)-2=0[/TEX]

[TEX]\Leftrightarrow 2cos^22x-cos2x-1=0[/TEX]

[TEX]\Leftrightarrow \left[cos2x-1=0\\cos2x+\frac{1}{2}=0[/TEX]

 

[silvery21]

[TEX]44/ tanx - sin2x - cos2x + 2(2cosx - \frac{1}{cosx}) = 0[/TEX]

đk........

\Leftrightarrow [TEX]\frac{sin x}{ co s x}- sin x co s x -cos2x + 2( \frac{2cos^2x -1}{cosx}) = 0[/TEX]

\Leftrightarrow [TEX]\frac{sin x}{ co s x}(1-2cos^2x) - cos2x + 2( \frac{cos2x -1}{cosx}) = 0[/TEX]

\Leftrightarrow[TEX] \- \frac{sin x}{ co s x} cos2x - cos2x+ 2( \frac{cos2x -1}{cosx}) = 0[/TEX]

\Leftrightarrow [TEX]\-cos2x( sin x + co s x)-2)=0[/TEX]

 

[rua_it]

[TEX]41/ cotx - 1 = \frac{cos2x}{1 + tanx} + sin^2x - \frac{1}{2}sin2x[/TEX]

[tex]cotx - 1 = \frac{cos2x}{1 + tanx} + sin^2x - \frac{1}{2}sin2x[/TEX]

[tex]\Rightarrow cotx-1=\frac{(cos^2x-sin^2x)}{1+\frac{sinx}{cosx}}+sin^2x - \frac{1}{2}sin2x[/TEX]

[tex]\Rightarrow \frac{cosx}{sinx}-1=\frac{cosx.(cos^2x-sin^2x)}{sinx+cosx}+sin^2x - \frac{1}{2}sin2x[/TEX]

$$\Rightarrow \frac{cosx-sinx}{sinx}=sin^2x+cos^2x-cosx.sinx-\frac{1}{2}sin2x$$

$$\Rightarrow cosx-sinx=sinx.(1-sin2x)$$

$$\Rightarrow cosx-sinx=sinx.(cosx-sinx)^2$$

 

[djbirurn9x]

$$cotx - 1 = \frac{cos2x}{1 + tanx} + sin^2x - \frac{1}{2}sin2x$$

$$\Rightarrow cotx-1=\frac{(cos^2x-sin^2x)}{1+\frac{sinx}{cosx}}+sin^2x - \frac{1}{2}sin2x$$

$$\Rightarrow \frac{cosx}{sinx}-1=\frac{cosx.(cos^2x-sin^2x)}{sinx+cosx}+sin^2x - \frac{1}{2}sin2x$$

$$\Rightarrow \frac{cosx-sinx}{sinx}=sin^2x+cos^2x-cosx.sinx-\frac{1}{2}sin2x$$

$$\Rightarrow cosx-sinx=sinx.(1-sin2x)$$

$$\Rightarrow cosx-sinx=sinx.(cosx-sinx)^2$$

Thiếu đk nha rua em
P/s : bài 43 mình biến đổi 1 tí mà trông khó ghê

 

[djbirurn9x]

đk........

\Leftrightarrow [TEX]\frac{sin x}{ co s x}- sin x co s x -cos2x + 2( \frac{2cos^2x -1}{cosx}) = 0[/TEX]

\Leftrightarrow [TEX]\frac{sin x}{ co s x}(1-2cos^2x) - cos2x + 2( \frac{cos2x -1}{cosx}) = 0[/TEX]

\Leftrightarrow[TEX] \- \frac{sin x}{ co s x} cos2x - cos2x+ 2( \frac{cos2x -1}{cosx}) = 0[/TEX]

\Leftrightarrow [TEX]\-cos2x( sin x + co s x)-2)=0[/TEX]

\Leftrightarrow [TEX]\frac{sin x}{ co s x}- sin x co s x -cos2x + 2( \frac{2cos^2x -1}{cosx}) = 0[/TEX]
thiếu số 2 [TEX](sin2x = 2sinxcosx)[/TEX] :|

 

[djbirurn9x]

chém nào

[TEX]46/ cos^4x + sin^6x = cos2x[/TEX]

[TEX]47/sin^2x + \frac{1}{4}sin^23x = sinxsin^23x[/TEX]

[TEX]48/sin^2x + sin^2y + sin^2(x+y) = \frac{9}{4}[/TEX]

[TEX]49/tan^2x + tan^2y + cot^2(x+y) = 1[/TEX]

[TEX]50/x^2 - 2xsinxy + 1 = 0[/TEX]

P/s: bài 43,45 chưa ai chém ah . Rua đến từ biển hả

 

[djbirurn9x]

Rua ơi vào đây giải hộ anh mấy bài BĐT :
http://diendan.hocmai.vn/showthread.php?p=1017850&posted=1#post1017850

 

[boon_angel_93]

[TEX]46/ cos^4x + sin^6x = cos2x[/TEX]

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

[TEX]\Leftrightarrow sin^4x+sin^6x=0[/TEX]

[TEX]\Leftrightarrow sin^4x.(1+sin^2x)=0[/TEX]xong

 

[boon_angel_93]

[TEX]49/tan^2x + tan^2y + cot^2(x+y) = 1[/TEX]

có [TEX]cotg(x+y)=\frac{1-tgxtgy}{tgx+tgy}[/TEX]

[TEX]\Leftrightarrow (tgx+tgy).cotg(x+y)=1-tgxtgy[/TEX]

[TEX]\Leftrightarrow cotg(x+y).tgx+cotg(x+y).tgy=1-tgxtgy[/TEX]

[TEX]\Leftrightarrow tgxtgy+tgx+cotg(x+y).tgy=1[/TEX](1)
thay bt (1) vào pt trên ta dc
[TEX]tan^2x + tan^2y + cot^2(x+y) =tgxtgy+tgx+cotg(x+y).tgy[/TEX]

[TEX]\Leftrightarrow 2tgxtgy+2tgxcot(x+y)+2tgy.cot(x+y)=2tg^2x+2tg^2y+2cot^2(x+y)[/TEX]

[TEX]\Leftrightarrow (tgx-tgy)^2+[tgx-cot(x+y)]^2+[tgy-cot(x+y)]^2=0[/TEX]

dấu '=' xảy ra \Leftrightarrow[TEX]tgx=tgy=cot(x+y)[/TEX] \Rightarrow[TEX]tg^2x=tg^2y=cot^2(x+y)[/TEX]

mà [TEX]tan^2x + tan^2y + cot^2(x+y) = 1[/TEX] \Rightarrow [TEX]tg^2x=tg^2y=cot^2(x+y)=\frac{1}{3}[/TEX]

\Rightarrow [TEX]tgx=tgy=cot(x+y)=\frac{+-\sqr{3}}{3}[/TEX]

\Rightarrow [TEX]tgx=\frac{\sqr{3}}{3}}[/TEX]\Leftrightarrow[TEX]x=\frac{\pi}{6}+k\pi[/TEX]

[TEX]tgx=\frac{-\sqr{3}}{3}[/TEX]\Leftrightarrow [TEX]x=\frac{-\pi}{6}+k\pi[/TEX]

tt với [TEX]tgy=\frac{\sqr{3}}{3}[/TEX]

và [TEX]tgy=\frac{-\sqr{3}}{3}[/TEX] ok

 

[silvery21]

[TEX]48/sin^2x + sin^2y + sin^2(x+y) = \frac{9}{4}[/TEX]

[TEX]VT= P= \frac{1-cos2x}{2}+ \frac{1-cos2y}{2}+sin^2(x+y)[/TEX]

[TEX]= 1- cos(x+y).cos(x-y)+1-cos^2(x+y)[/TEX]

[TEX]\Leftrightarrow cos^2(x+y)+ cos(x+y).cos(x-y) +P-2=0[/TEX] có nghiệm .

[TEX]\Leftrightarrow \Delta=cos^2(x-y)- 4P+8\ge 0[/TEX]

[TEX]\Rightarrow P\le \frac{9}{4} [/TEX]

[TEX]sin^2x + sin^2y +sin^2(x+y) =\frac{9}{4}[/TEX]

[TEX]\Leftrightarrow \left{cos(x-y)=\pm 1\\cos(x+y)=\pm \frac{1}{2}[/TEX]

 

[silvery21]

[TEX]50/x^2 - 2xsinxy + 1 = 0[/TEX]

[TEX]\Delta'= sin ^2 xy -1 = - cos^2 xy \leq 0[/TEX]

=> pt có ngh \Leftrightarrow[TEX] \Delta'=0 \Leftrightarrow cos x y =0[/TEX]

\Leftrightarrow[TEX]\Leftrightarrow \left{cos(xy)=0\\x=sin xy[/TEX]

\Leftrightarrow [TEX]x=+-1 ; y= \pi/2 + 2 k\pi[/TEX]

 

[silvery21]

[TEX]47/sin^2x + \frac{1}{4}sin^23x = sinxsin^23x[/TEX]

[TEX]\Leftrightarrow sin^2x+\frac{sin^23x}{4}=sinxsin^23 x[/TEX]

[TEX]\Leftrightarrow (sinx-\frac{sin^23x}{2})^2+\frac{sin^23x}{4}(1-sin^23x)=0[/TEX]

[TEX]\Leftrightarrow (sinx-\frac{sin^23x}{2})^2+\frac{sin^26x}{16}=0[/TEX]
đến đây là OK.........

 

[silvery21]

[TEX]43/ \frac{3sin3x - \sqrt{3}cos9x - 1}{sin2x + sinx} = 4sin^23x(2cosx - 1)[/TEX]

[TEX]45/ (1 + 2cos3x)sinx + sin2x = 2sin^2(2x + \frac{\pi}{4})[/TEX]

b 43 ko ra ; anh xem lại đề

45 ; \Leftrightarrow[TEX] sin x + 2cos3xsinx + sin2x = 1-cos ( 4x+\pi/2)[/TEX]

\Leftrightarrow[TEX]sin x + sin 4x - sin 2x + sin 2x= 1 + sin 4x[/TEX]

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

 

[quyenuy0241]

[TEX][TEX]43/ \frac{3sin3x - \sqrt{3}cos9x - 1}{sin2x + sinx} = 4sin^23x(2cosx - 1)[/TEX]

DKXD tự làm nhá:
$$\frac{3sin3x-\sqrt{3}(4cos^33x-3cos3x-1)}{sinx(2cosx+1)}=4sin^23x(2cosx - 1)$$

$$3sin3x+3\sqrt{3}cos3x-\sqrt{3}.4cos^33x-1=sin^23x.sinx.(4cos^2x-1)$$

$$\Leftrightarrow 3sin3x+3\sqrt{3}cos3x-\sqrt{3}.4cos^33x-1=4sin^23x(3sinx-4sin^3x)$$

$$\Leftrightarrow 3sin3x+3\sqrt{3}cos3x-\sqrt{3}.4cos^33x-1=4sin^33x$$

$$\Leftrightarrow sin9x-\sqrt{3}cos9x=1 \Leftrightarrow sin(9x-\frac{\pi}{3})=1$$