Toán 11 Phương trình lượng giác.

[silvery21]

[TEX]13/ cos3xcos^3x - sin^3x = \frac{2 + 2\sqrt{3}}{8}[/TEX]

[

[TEX]\Leftrightarrow cos3x.(cos3x+3cosx)/4 - sin3x(-sin3x+3sinx)/4 =(2+2 \sqrt{3})/8[/TEX]

[TEX]\Leftrightarrow 1/4( cos(^2)3x+sin(^2)3x )+ 3/4 (cosx.cos3x - sinx.sin3x) =(2+2 \sqrt{3})/8[/TEX]

[TEX]\Leftrightarrow 1/4 +3/4 cos4x=(2+2 \sqrt{3})/8[/TEX]

[TEX]\Rightarrow cos 4x =1/\sqrt{3} \Rightarrow..............[/TEX]

 

[silvery21]

[TEX]11/ sinxcos4x - sin^22x = 4sin^2(\frac{\pi}{4} - \frac{x}{2}) - \frac{7}{2}[/TEX]

[TEX]12/ 2sin^22x + sin7x -1 = sinx[/TEX]

[TEX]13/ cos3xcos^3x - sin^3x = \frac{2 + 3\sqrt{2}}{8}[/TEX]

[TEX]14/ (2sinx +1)(3cos4x + 2sinx - 4) + 4cos^2x = 3[/TEX]

[TEX]15/ tan^2x - tanxtan3x = 2[/TEX]

Ủng hộ nào, thax cái nha

mai mình giải cho

h kovik đc;gõ mãi mới đc mi' chữ|

 

[silvery21]

[TEX]14/ (2sinx +1)(3cos4x + 2sinx - 4) + 4cos^2x = 3[/TEX]

Ta có : [TEX] 4cos^2x-3 =1-4sin^2x=(1+2sinx)(1-2sinx)[/TEX] (1)

=>

[TEX](2sinx+1)(3cos4x+2sinx-4+1-2sinx)=0[/TEX]

\Leftrightarrow[TEX](2sinx+1)(3cos4x-3)=0[/TEX]

[TEX] OK...................[/TEX]

 

[silvery21]

[TEX]15/ tan^2x - tanxtan3x = 2[/TEX]

Ủng hộ nào, thax cái nha

\Leftrightarrow [TEX]tan^2x - tanx\frac{3tgx - tg^3x}{1-3tg^2x} =2[/TEX]

\Leftrightarrow [TEX](tg^2x-1)^2 =0 ................................OK[/TEX]

trưa về giải ......đi học.............

 

[rua_it]

[TEX]11/ sinxcos4x - sin^22x = 4sin^2(\frac{\pi}{4} - \frac{x}{2}) - \frac{7}{2}[/TEX]

Ủng hộ nào, thax cái nha

[tex]sinxcos4x - sin^22x = 4sin^2(\frac{\pi}{4} - \frac{x}{2}) - \frac{7}{2}[/tex]

[tex]\Rightarrow sinx.cosx+\frac{cos4x}{2}-\frac{1}{2}+2sinx+\frac{3}{2}=2-2.cos(\frac{\pi}{2}-x)-\frac{7}{2}[/tex]

[tex]\Rightarrow sinx.cosx-\frac{1}{2}.(1-cos4x)+2sinx+\frac{3}{2}=0[/tex]

[tex]\Rightarrow sinxcos4x+2sinx+\frac{1}{2}cos4x+1=0[/tex]

[tex]\Rightarrow (sinx+\frac{1}{2}).(cos4x+2)=0[/tex]

[tex]\Rightarrow \left[\begin{sinx+\frac{1}{2}=0}\\{cos4x+2=0[/tex]

[tex]\Rightarrow \left[\begin{sinx=-\frac{1}{2}}\\{cos4x=-2}[/tex]

ok.

 

[botvit]

[TEX]11/ sinxcos4x - sin^22x = 4sin^2(\frac{\pi}{4} - \frac{x}{2}) - \frac{7}{2}[/TEX]

[TEX]12/ 2sin^22x + sin7x -1 = sinx[/TEX]



Ủng hộ nào, thax cái nha

cau 12
PT\Leftrightarrow[TEX]sin7x-sinx=1-2sin^22x[/TEX]
\Leftrightarrow[TEX]2cos4x.cos3x=cos4x[/TEX]
\Leftrightarrow[TEX]cos4x(2cos3x-1)=0[/TEX]

 

[djbirurn9x]

\Leftrightarrow [TEX]tan^2x - tanx\frac{3tgx - tg^3x}{1-3tg^2x} =2[/TEX]

\Leftrightarrow [TEX](tg^2x-1)^2 =0 ................................OK[/TEX]

trưa về giải ......đi học.............

Giải đúng nhưng bài nào có [TEX]tan, cot[/TEX] hay có mẫu thì nên ghi điều kiện là [TEX]x \not = .......[/TEX]
, rồi so đk nhận / loại nghiệm

 

[botvit]

[TEX]15/ tan^2x - tanxtan3x = 2[/TEX]

Ủng hộ nào, thax cái nha

1 cach #.............................................
dkcosx #0;cos4x+cos2x#0
PT\Leftrightarrow
PT\Leftrightarrow [TEX]\frac{sin^2x}{cos^2x}-\frac{sinx.sin3x}{cosx.cos3x}=2[/TEX]
ta co [TEX]sinxsin3x=-\frac{1}{2}(cos4x+cos2x)[/TEX]
[TEX]cosxcos3x=\frac{1}{2}(cos4x+cos2x)[/TEX]
\Rightarrow [TEX]\frac{sin^2x}{cos^2x}+1=2[/tex]

 

[djbirurn9x]

[TEX]16/ sin2x(cotx + tan2x) = 4cos^2x[/TEX]

[TEX]17/ \frac{cot^2x - tan^2x}{cos2x} = 16(1 + cos4x)[/TEX]

[TEX]18/ sin^4x + cos^4x = \frac{7}{8}cot(x + \frac{\pi}{3})cot(\frac{\pi}{6} - x)[/TEX]

[TEX]19/ 2tanx + cot2x = 2sin2x + \frac{1}{sin2x}[/TEX]

[TEX]20/ sin(\frac{3\pi}{10} - \frac{x}{2}) = \frac{1}{2}sin(\frac{\pi}{10} + \frac{3x}{2})[/TEX]

Thax ủng hộ nha , bài nào có điều kiện nhớ ghi ra

 

[quyenuy0241]

[tex]cos3x.cos^3x-sin3xsin^3x=\frac{2+2\sqrt{3}}{8}[/tex]
....................................

 

[botvit]

[TEX]16/ sin2x(cotx + tan2x) = 4cos^2x[/TEX]

[TEX]17/ \frac{cot^2x - tan^2x}{cos2x} = 16(1 + cos4x)[/TEX]

[TEX]18/ sin^4x + cos^4x = \frac{7}{8}cot(x + \frac{\pi}{3})cot(\frac{\pi}{6} - x)[/TEX]

[TEX]19/ 2tanx + cot2x = 2sin2x + \frac{1}{sin2x}[/TEX]

[TEX]20/ sin(\frac{3\pi}{10} - \frac{x}{2}) = \frac{1}{2}sin(\frac{\pi}{10} + \frac{3x}{2})[/TEX]

Thax ủng hộ nha , bài nào có điều kiện nhớ ghi ra

16. dk tu lam
.................................
PT\Leftrightarrow [TEX]sin2x(\frac{cosx}{sinx}+\frac{cos2x}{sin2x})=4cos^2x[/TEX]
\Leftrightarrow[TEX] \frac{sin2xcosx}{cos2x}=4cos^2x[/TEX]
\Leftrightarrow[TEX]cosx(sin2x-cos2x4cosx)=0[/TEX]

 

[ngomaithuy93]

[tex]cos3x.cos^3x-sin3xsin^3x=\frac{2+2\sqrt{3}}{8}[/tex]
....................................

[TEX] cos3xcos^3x-sin3xsin^3x=\frac{2+2\sqrt{3}}{8}[/TEX]
\Leftrightarrow [TEX] -3cos^4x+4cos^6x-3sin^4x+4sin^6x=\frac{2+2\sqrt{3}}{8}[/TEX]
\Leftrightarrow [TEX] 6sin^2xcos^2x-12sin^4xcos^2x-12sin^2xcos^4x=\frac{2+2\sqrt{3}}{8}[/TEX]
\Leftrightarrow [TEX] 6sin^2xcos^2x=\frac{-2-2\sqrt{3}}{8}[/TEX]
Vô nghiệm! Ko biết đúng sai ra làm sao nữa!

 

[botvit]

[tex]cos3x.cos^3x-sin3xsin^3x=\frac{2+2\sqrt{3}}{8}[/tex]
....................................

PT\Leftrightarrow[TEX]8cos3xcos^3x-8sin3xsin^3x=2+2\sqrt{3}[/TEX]
[TEX]\Leftrightarrow 2cos3x(cos3x+3cosx)-2sin3x(3sinx-sin3x)=2+2\sqrt{3}[/TEX]
\Leftrightarrow[TEX] 2cos^23x+6cos3x.cosx+2sin^23x-6sin3x.sinx=2+2\sqrt{3}[/TEX]
\Leftrightarrow[TEX]2(cos^23x+sin^23x)+3(cos4x+cos2x)+3(cos4x-cos2x)=2+2\sqrt{3}[/TEX]
\Leftrightarrow[TEX]2+6cos4x=2+2\sqrt{3}[/TEX]
\Leftrightarrow[TEX]cos4x=\frac{1}{\sqrt[]{3}}[/TEX]

 

[ngomaithuy93]

[TEX] [TEX]18/ sin^4x + cos^4x = \frac{7}{8}cot(x + \frac{\pi}{3})cot(\frac{\pi}{6} - x)[/TEX]

[TEX] sin^4x+cos^4x=\frac{7}{8}cot(x+\frac{\pi}{3})cot(.\frac{\pi}{6}-x)[/TEX]
\Leftrightarrow [TEX] -2sin^2xcos^2x=\frac{7}{8}.\frac{cos(2x-\frac{\pi}{6})+cos.\frac{\pi}{2}}{cos(2x-\frac{\pi}{6})+cos.\frac{\pi}{2}}[/TEX]
\Leftrightarrow [TEX] sin^2xcos^2x=\frac{-7}{16}[/TEX]
Lại vô nghiệm nữa sao?
Hay tớ giải sai nhỉ?

 

[quyenuy0241]

[TEX]18/ sin^4x + cos^4x = \frac{7}{8}cot(x + \frac{\pi}{3})cot(\frac{\pi}{6} - x)[/TEX]

DK tự làm nhé......
Để ý rằng [tex]cot(x + \frac{\pi}{3})=tan(\frac{\pi}{6}-x)[/tex]
[tex]Pt \Leftrightarrow 8(cos^4x+cos^4x)=7 \Leftrightarrow 8(1-2sin^2xcos^2x)=7 \Leftrightarrow 4sin^22x=1 \Leftrightarrow ........[/tex]

 

[botvit]

[TEX]18/ sin^4x + cos^4x = \frac{7}{8}cot(x + \frac{\pi}{3})cot(\frac{\pi}{6} - x)[/TEX]

Thax ủng hộ nha , bài nào có điều kiện nhớ ghi ra

dk tu lam pt
....................................................
ta co:
[TEX]cot(x+\frac{pi}{3}).cot(\frac{pi}{6}-x)=\frac{\frac{1}{2}.cos.\frac{pi}{2}.cos(2x+\frac{pi}{6})}{\frac{1}{2}.cos.\frac{pi}{2}.cos(2x+\frac{pi}{6})}=1[/TEX]

\Leftrightarrow[TEX]8(sin^4x+cos^4x)=7[/TEX]
\Leftrightarrow[TEX]8(1-\frac{1}{2}sin^22x)=7[/TEX]

 

[botvit]

[TEX]17/ \frac{cot^2x - tan^2x}{cos2x} = 16(1 + cos4x)[/TEX]

Thax ủng hộ nha , bài nào có điều kiện nhớ ghi ra

dk tu lam
.........................................................
ta co [TEX]\frac{cot^2x - tan^2x}{cos2x}=\frac{\frac{cos^2x}{sin^2x}-\frac{sin^2x}{cos^2x}}{cos2x}=\frac{\frac{cos^4x-sin^4x}{sin^2xcos^2x}}{cos2x}=\frac{cos^22x}{sin^2xcos^2x}[/TEX]
PT\Leftrightarrow [TEX]cos^22x=16sin^2xcos^2x(1+cos4x)[/TEX]
\Leftrightarrow[TEX]cos^22x=4sin^22x(1+cos4x)[/TEX]
\Leftrightarrow[TEX]2cos^22x=8sin^22x(1+cos4x)[/TEX]
\Leftrightarrow[TEX]cos4x+1=8sin^22x(1+cos4x)[/TEX]

 

[silvery21]

[TEX] cos3xcos^3x-sin3xsin^3x=\frac{2+2\sqrt{3}}{8}[/TEX]
\Leftrightarrow [TEX] -3cos^4x+4cos^6x-3sin^4x+4sin^6x=\frac{2+2\sqrt{3}}{8}[/TEX]
\Leftrightarrow [TEX] 6sin^2xcos^2x-12sin^4xcos^2x-12sin^2xcos^4x=\frac{2+2\sqrt{3}}{8}[/TEX]
\Leftrightarrow [TEX] 6sin^2xcos^2x=\frac{-2-2\sqrt{3}}{8}[/TEX]
Vô nghiệm! Ko biết đúng sai ra làm sao nữa!

cậu nhìnkĩ ha ;t giải ròi


[TEX]\Leftrightarrow cos3x.(cos3x+3cosx)/4 - sin3x(-sin3x+3sinx)/4 =(2+2 \sqrt{3})/8[/TEX]

[TEX]\Leftrightarrow 1/4( cos(^2)3x+sin(^2)3x )+ 3/4 (cosx.cos3x - sinx.sin3x) =(2+2 \sqrt{3})/8[/TEX]

[TEX]\Leftrightarrow 1/4 +3/4 cos4x=(2+2 \sqrt{3})/8[/TEX]

[TEX]\Rightarrow cos 4x =1/\sqrt{3} \Rightarrow..............[/TEX]

 

[rua_it]

[TEX] [TEX]19/ 2tanx + cot2x = 2sin2x + \frac{1}{sin2x}[/TEX]


Thax ủng hộ nha , bài nào có điều kiện nhớ ghi ra

[tex] 2tanx + cot2x = 2sin2x + \frac{1}{sin2x}[/TEX]

[tex]\Rightarrow 2.\frac{sinx}{cosx}+\frac{cos2x}{sin2x}-2sin2x-\frac{1}{sin2x}=0[/tex]

[tex]\Rightarrow 4sin^2x-8sin^2xcos^2x-2sin^2x=0[/tex]

[tex]\Rightarrow 4sin^2x+cos2x-2sin^22x-1=0[/tex]

[tex]\Rightarrow 2sin^2x.(1-4cos^2x)=0[/tex]

 

[quyenuy0241]

cậu nhìnkĩ ha ;t giải ròi


[TEX]\Leftrightarrow cos3x.(cos3x+3cosx)/4 - sin3x(-sin3x+3sinx)/4 =(2+3 \sqrt{2})/8[/TEX]

[TEX]\Leftrightarrow 1/4( cos(^2)3x+sin(^2)3x )+ 3/4 (cosx.cos3x - sinx.sin3x) =(2+3 \sqrt{2})/8[/TEX]

[TEX]\Leftrightarrow 1/4 +3/4 cos4x=(2+3 \sqrt{2})/8[/TEX]

[TEX]\Rightarrow cos 4x =1/\sqrt{2} \Rightarrow..............[/TEX]

Có nhầm không bạn !!!Mình đổi đề roài !!!
Đề là[tex] 2\sqrt{3}[/tex] mà