giải phương trình

N

nguyenbahiep1

1/cos 10x-cos 8x-cos 6x+1=0
2/8cos^4(x+pi/4)+sin4x=2.\frac{1-tan^2 x}{1+tan^2x}
3/1-2căn3.sin x+3sin^2 x=cos^2 2x
4/8cos x+6sin x-cos 2x-7=0
5/sin(3pi/5+2x)=2sin^2(pi/5-x)
6/1+sinx+cosx+sin2x+cos2x=0

câu 6

[TEX]1 + sin 2x + sinx +cosx + cos^2x -sin^2x = (sinx+cosx)^2 + (sin x + cosx) + (cosx +sin x) (cosx -sin x) = 0 \\ sin x +cosx = 0 \Rightarrow tan x = -1 \\ sin x + cosx + 1 +cosx -sin x = 0 \Rightarrow cosx = -\frac{1}{2}[/TEX]
 
G

gau_gau_gau_00

giải pt

câu 6
1+sin(x) +cox(x) +sin(2x) +cox(2x)=0
<=>1+sin(x) +cox(x) +2sin(x)cox(x) +2cos^2(x)-1=0
<=>sin(2cos +1) + cos(2cos +1)=0
<=>(2cos +1)(sin+cos)=0
.
cos(x)=-1/2 =>x=+-(3/2)pi +k2pi
.
sin=-cos
=>tan(x)=-1
=>x=pi/4 +kpi
 
N

newstarinsky

$3)1-2\sqrt{3}sinx+3sin^2x=1-sin^22x\\
\Leftrightarrow 3sin^2x-2\sqrt{3}sinx+4sin^2x.cos^2x=0\\
\Leftrightarrow sinx[3sinx-2\sqrt{3}+4sinx(1-sin^2x)]=0\\
\Leftrightarrow sinx[-4sin^3x+7sinx-2\sqrt{3}]=0$

$1) cos10x-cos6x+1-cos8x=0\\
\Leftrightarrow -2sin8x.sin2x+2sin^24x=0\\
\Leftrightarrow -2sin4x.cos4x.sin2x+2sin4x.sin2x.cos2x=0\\
\Leftrightarrow sin4x.sin2x(cos2x-cos4x)=0\\
\Leftrightarrow sin4x.sin2x.sin3x.sinx=0$
 
Last edited by a moderator:
L

locxoaymgk

Bài 1:
[TEX]cos10x - cos8x - cos6x + 1 = 0[/TEX]

[TEX]\Leftrightarrow cos 10x - cos 6x + 1 - cos 8x = 0[/TEX]

[TEX]\Leftrightarrow -2.sin 8x.sin 2x + 2.sin^2(4x) = 0[/TEX]

[TEX]\Leftrightarrow -2.sin4x.cos 4x.sin 2x + sin^2(4x) = 0[/TEX]

[TEX] \Leftrightarrow sin 4x( sin 4x- 2cos 4x.sin 2x) = 0[/TEX]

[TEX]\Leftrightarrow sin 4x ( 2.sin 2x . cos 2x - 2.cos 4x .sin 2x) = 0[/TEX]

[TEX]\Leftrightarrow sin 4x .sin 2x. (cos 2x - cos 4x) = 0[/TEX]

[TEX]\Leftrightarrow sin 4x . sin 2x . ( - 2.cos^2(2x) + cos 2x + 1) = 0[/TEX]
 
Top Bottom