P
S
silvery21
câu 2 nak`
pt\Leftrightarrow[TEX]2(cosx-cos28xsinx)=3\sqrt{2}-\sqrt{2}sin10x[/TEX]
[TEX]VT=2(cosx-cos28xsinx)[/TEX] \leq [TEX]2\sqrt{1+cos^228x)} \leq 2\sqrt{2}[/TEX]
[TEX]Vp \geq 3\sqrt{2}- \sqrt{2}=2\sqrt{2}[/TEX]
dâu' =
\Leftrightarrow [TEX]cos^228x=1[/TEX]
và [TEX] cosxcos28x=-sinx[/TEX]
[TEX]sin10x=1[/TEX]
\Leftrightarrow [TEX]x=pi/4 + kpi[/TEX]
Last edited by a moderator:
N
ngomaithuy93
[TEX]sin4x-cos4x=1+4\sqrt{2}sin(x-\frac{\pi}{4})[/TEX]
\Leftrightarrow[TEX]2sin2x(cos2x-sin2x)=4(sinx-cosx)[/TEX]
\Leftrightarrow[TEX]4sinxcosx(-4sin^2xcos^2x-2sinxcosx+1)=4(sinx-cosx)[/TEX]
Đặt [TEX]sinx-cosx=t \Rightarrow sinxcosx=\frac{1-t^2}{2}[/TEX]
C
chienam1c
>-
V
vitcon10
D
doremon.
làm tiếp nhé
1.3cos2x+4sin2x-9cosx-4sinx+6=0
2.cosx+cos3x+cos5x=-2
3.[tex]tan^2x+cot^2x=sin^3y[/tex]
4.[tex]\frac{cos^{10}x+sin^{10}x}{4}=\frac{sin^6x+cos^6x}{4cos^22x+sin^22x}[/tex]
1.3cos2x+4sin2x-9cosx-4sinx+6=0
2.cosx+cos3x+cos5x=-2
3.[tex]tan^2x+cot^2x=sin^3y[/tex]
4.[tex]\frac{cos^{10}x+sin^{10}x}{4}=\frac{sin^6x+cos^6x}{4cos^22x+sin^22x}[/tex]
N
ngomaithuy93
[TEX]Pt \Leftrightarrow 6cos^2x-9cosx+3+4sinx(2cosx-1)=0[/TEX]
\Leftrightarrow[TEX] (3cosx-3)(2cosx-1)+4sinx(2cosx-1)=0[/TEX]
\Leftrightarrow [TEX](2cosx-1)(3cosx+4sinx-3)=0[/TEX]