P
police92
[TEX]sin^4(x)+sin^4([TEX]\frac{x}\frac{2}[/TEX]+[TEXT]\frac{\pi}\frac{8}[/TEX])+C0S^4(x)=0,5sin^2(2x)[/TEX]
L
lantrinh93
N
nhoc_maruko9x
[TEX]\Leftrightarrow 2cos3x(2-4sin^2x+1)=1[/TEX])
giải pt lượng giác :
[TEX]2cos3x.(2.cos2x+1)=1[/TEX]
.......................
[TEX]\Leftrightarrow 2cos3x(3-4sin^2x)=1[/TEX]
[TEX]\Leftrightarrow 2cos3x(3sinx-4sin^3x)=sinx[/TEX] (Xét sinx = 0 trước
)[TEX]\Leftrightarrow 2cos3x.sin3x=sinx[/TEX]
[TEX]\Leftrightarrow sin6x=sinx[/TEX]
Đến đây thui nhá!
L
lucky_star93
[TEX]sin4x +cos3x +cosx = 4sinx+2[/TEX]
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0
0vietsang0
tiếp vài bài nữa này :
[TEX]a) \sqrt{3}sin3x + 4cos^3x = 1 + 3sin(\frac{\pi}{2}-x)\\ b) sin^3xcos3x + cos^3xsin3x= sin^34x \\ c) \frac{1-2cos^2x}{sinxcosx} + 2tan2x + cot^34x =3[/TEX]
[TEX]a) \sqrt{3}sin3x + 4cos^3x = 1 + 3sin(\frac{\pi}{2}-x)\\ b) sin^3xcos3x + cos^3xsin3x= sin^34x \\ c) \frac{1-2cos^2x}{sinxcosx} + 2tan2x + cot^34x =3[/TEX]
N
nhocngo976
\Leftrightarrow[TEX]\sqrt{3}sin3x+cos3x=1[/TEX]
\Leftrightarrow[TEX]2sin(3x+\frac{\pi}{6})=1[/TEX]
N
nhocngo976
N
nhocngo976
DK...
\Rightarrow[TEX]\frac{-2cos2x}{sin2x}+2tan2x+cot^34x=3[/TEX]
\Leftrightarrow[TEX] -2cot2x+2tan2x+ cot^34x=3[/TEX]
tự xử
)có thể đặt tan2x=t ( khó )
N
ngomaithuy93
R
roses_123
[TEX] tan(2x-\frac{\pi}{4})tan(2x+\frac{\pi}{4})=\frac{4cos^22x}{tanx-cotx} [/TEX]
Sai rồi!
ở đâu thế
ĐK:
xét [TEX]tan(2x-\frac{\pi}{4})tan(2x+\frac{\pi}{4})=-1[/TEX]
ft [TEX]\Leftrightarrow \frac{4cos^22x}{tanx-cotx} =-1[/TEX]
T
tranhuukha
O
oaivy93somic
L
lamtrang0708
dùng tanx= sinx/cosx
=> pt trở thành
[tex] cosx - 2sinx.cos^2 x= 2sin^2x.cosx-sinx[/tex] (quy đồng)
=> (cosx-sinx)(1-sin2x)=0
....
bạn tự làm tiếp nhé
[tex]2sin^2(x - \frac{\pi}{4}) = -sin2x [/tex] (chắc vậy
)
D
doremon_1190
S
siengnangnhe
L
lamtrang0708
dễ thấy 2cosx-3<0 do đó :
[tex] \begin{array}{l} \Leftrightarrow \frac{{\sqrt 3 (c{\rm{o}}{{\rm{s}}^2}x + \cos x - 2)}}{{2\cos x - 3}} = {\{\rm s}\nolimits} {\rm{inx}}\\\\ {\sin ^2}x + c{\rm{o}}{{\rm{s}}^2}x = 1 \Rightarrow \frac{{3{{(c{\rm{o}}{{\rm{s}}^2}x + \cos x - 2)}^2}}}{{{{(2\cos x - 3)}^2}}} + c{\rm{o}}{{\rm{s}}^2}x = 1\\\\ \Leftrightarrow 7c{\rm{o}}{{\rm{s}}^4}x - 6c{\rm{o}}{{\rm{s}}^3}x - 4c{\rm{o}}{{\rm{s}}^2}x + 3 = 0\\\\ \Leftrightarrow (\cos x - 1)(7c{\rm{o}}{{\rm{s}}^3}x + c{\rm{o}}{{\rm{s}}^2}x - 3c{\rm{os}}x - 3) = 0\\\\ \Leftrightarrow \left[ {\begin{array}{*{}{}} {\cos x = 1}\\\\ {7c{\rm{o}}{{\rm{s}}^3}x + c{\rm{o}}{{\rm{s}}^2}x - 3c{\rm{os}}x - 3 = 0} \end{array}} \right. \end{array}[/tex]
[tex] \begin{array}{l} \Leftrightarrow \frac{{\sqrt 3 (c{\rm{o}}{{\rm{s}}^2}x + \cos x - 2)}}{{2\cos x - 3}} = {\{\rm s}\nolimits} {\rm{inx}}\\\\ {\sin ^2}x + c{\rm{o}}{{\rm{s}}^2}x = 1 \Rightarrow \frac{{3{{(c{\rm{o}}{{\rm{s}}^2}x + \cos x - 2)}^2}}}{{{{(2\cos x - 3)}^2}}} + c{\rm{o}}{{\rm{s}}^2}x = 1\\\\ \Leftrightarrow 7c{\rm{o}}{{\rm{s}}^4}x - 6c{\rm{o}}{{\rm{s}}^3}x - 4c{\rm{o}}{{\rm{s}}^2}x + 3 = 0\\\\ \Leftrightarrow (\cos x - 1)(7c{\rm{o}}{{\rm{s}}^3}x + c{\rm{o}}{{\rm{s}}^2}x - 3c{\rm{os}}x - 3) = 0\\\\ \Leftrightarrow \left[ {\begin{array}{*{}{}} {\cos x = 1}\\\\ {7c{\rm{o}}{{\rm{s}}^3}x + c{\rm{o}}{{\rm{s}}^2}x - 3c{\rm{os}}x - 3 = 0} \end{array}} \right. \end{array}[/tex]
H
huu_thuong
V
vanglai
nếu sỉn 3x hem ra được đâu!!!- dùng Fx-570 thủ nghiệm lẻ à!!!
[TEX]sin4x +4cos^3x-2cosx=4sin x +2 [/TEX]
[TEX]4sin x( co sx cos2x-1)+2( cosx cos2x -1)=0[/TEX]
=> DoNe
V
vanculete
Q
quynhsuu_nghe_93
m(sinx+cosx+1)=1+2sinx.cosx tim m de pt co nghiem thuoc [0;pi/2]
baj tap naj co the lam nhu sau:
sinx+cosx=a voi x thoc [0;pi/2] \Rightarrow a thuoc [-1;1] (dung duong tron luong giac)
\Rightarrow m(a+1)=a^2
xet a=-1 \Rightarrow thoa man
xet voi a ≠ -1 \Rightarrow m=a^2/(a+1) voi a thuoc (-1;1]
su dung bang dong bien nghjch bien \Rightarrow m thuoc [o;+\infty ]
baj tap naj co the lam nhu sau:
sinx+cosx=a voi x thoc [0;pi/2] \Rightarrow a thuoc [-1;1] (dung duong tron luong giac)
\Rightarrow m(a+1)=a^2
xet a=-1 \Rightarrow thoa man
xet voi a ≠ -1 \Rightarrow m=a^2/(a+1) voi a thuoc (-1;1]
su dung bang dong bien nghjch bien \Rightarrow m thuoc [o;+\infty ]