[ help me]PT lượng giác cực hay

[jerusalem]

g) [TEX]sin^{8}x+cos^{8}x=2(sin^{10}x+cos^{10}x)+\frac{5}{4}cos2x[/TEX]

\Leftrightarrow[TEX]sin^8x+cos^8x=2[sin^8x(1-cos^2x)+cos^8(1-sin^2x)]+5/4cos2x[/TEX]
\Leftrightarrow[TEX]cos^8x.cos2x-sin^8x.cos2x+5/4cos2x=O[/TEX]
\Leftrightarrow[TEX]cos2x(cos^8x-sin^8x+5/4)=O[/TEX]
\Leftrightarrow[TEX]cos2x[(2cos^2x-1)(2cos^4x-2cos^2x+1)+5/4]=O[/TEX]
\Leftrightarrow[TEX]cos2x(2cos^6x-3cos^4x+2cos^2x+1/8)=O[/TEX]
\Leftrightarrow[TEX]\left[\begin{cos2x=O}\\{ 2cos^6x-3cos^4x+2cos^2x+1/8= O}[/TEX]
đặt [TEX]cos^2x=t[/TEX].giải như bt |

h) [TEX]3tan^{3}x-tanx+\frac{3(1+sinx)}{cos^{2}x}-8cos^{2}(\frac{\pi }{4}-\frac{x}{2})=0[/TEX]

[TEX]tan x(3tan^2x-1)+[\frac{3}{1-sinx}-4(1+sin x)]=O[/TEX]
\Leftrightarrow[TEX]tan x(\frac{4sin^2x-1}{1-sin^2x})+\frac{4sin^2x-1}{1-sin x}=O[/TEX]
\Leftrightarrow[TEX]\frac{4sin^2x-1}{1-sin x}(\frac{tan x}{1+sin x}+1)=O[/TEX]
\Leftrightarrow[TEX]\frac{4sin^2x-1}{1-sin x}(\frac{tan x+sin x+1}{1+sin x})=O[/TEX]
\Leftrightarrow[TEX]\left[\begin{4sin^2x-1=O}\\{tan x+sin x+1 =O ($) } [/TEX]

giải ($)
[TEX]tan x+sin^2 x/2+cos^2 x/2+2sin x .cos x=O[/TEX]
\Leftrightarrow[TEX]( sin x/2+cos x/2)^2=-tan x[/TEX]
\Leftrightarrow[TEX]( sin x/2+cos x/2)^2=\frac{-2sin x/2.cos x/2}{(sin x/2+cos x/2)(sin x/2-cos x/2)}[/TEX]
\Leftrightarrow[TEX]( sin x/2+cos x/2)^3=-2sin x/2.cos^2 x/2-2sin^2 x/2.cos x/2[/TEX]
\Leftrightarrow[TEX](sin x/2+cos x/2)(1+2sinx)=O[/TEX]

mệt quá|

 

[hetientieu_nguoiyeucungban]

[TEX] 1) sin^{3}x.sin3x+cos^{3}x.cos3x=\frac{\sqrt{2}}{4}[/TEX]

[TEX]2)sin^{3}x.sin3x+cos^{3}x.cos3x=cos^{3}4x[/TEX]

[TEX]3.-sin^{3}x.sin3x+cos^{3}x.cos3x=cos^{3}4x+\frac{1}{4}[/TEX]

[TEX]4.sin^{3}2x.cos6x+cos^{3}2x.sin6x=\frac{3}{8}[/TEX]

[TEX]5(2sinx+1)(3cos4x+2sinx-4)+4cos^{2}x=3.[/TEX]

 

[sontg12]

bài 1
[TEX]sinx^3sin3x +cosx^3cos3x=\frac{\sqrt{2}}{4}[/TEX]
[TEX]\frac{1}{2}sinx^2(cos2x-cos4x)+\frac{1}{2}cosx^2(cos2x+cos4x)=\frac{\sqrt{2}}{4}[/TEX]
[TEX] -2sinx^2cos4x+cos2x=\frac{\sqrt{2}}{2} [/TEX]
[TEX]3cos2x+cos6x=\sqrt{2} ta co4cosx^3=\sqrt{2}[/TEX]
[TEX]cosx=\frac{\sqrt{2}}{2}[/TEX]
ri dùng công thức hạ bậc về pt bậc ba nha

 

[sontg12]

mấy bài kia tương tự nha
bài cuối nè
[TEX](2sinx+1)(3cos4x+2sinx-4)+4cos^2x=3[/TEX]
[TEX]6sinxcos4x+4sin^2x-6sinx+3cos4x-4=3[/TEX]
[TEX]6sinxcos4x-6sinx+3cos4x-3=0[/TEX]
[TEX]6sinx(cos4x-1)+3(cos4x-1)=0[/TEX]
đên đây là ra rùi ha

 

[connguoivietnam]

[TEX]sin^3xsin3x+cos^3xcos3x=cos^34x[/TEX]

[TEX]sin^2x(cos2x-cos4x)+cos^2x(cos4x+cos2x)=2cos^34x[/TEX]

[TEX]sin^2xcos2x-sin^2xcos4x+cos^2xcos4x+cos^2xcos2x=2cos^34x[/TEX]

[TEX]cos2x+cos4x(cos^2x-sin^2x)=2cos^34x[/TEX]

[TEX]cos2x+cos4xcos2x=2cos^34x[/TEX]

[TEX]cos2x(1+cos4x)=2(2cos^22x-1)^3[/TEX]

[TEX]2cos^32x=2(8cos^62x-12cos^42x+6cos^22x-1)[/TEX]

[TEX]cos^32x=8cos^62x-12cos^42x+6cos^22x-1[/TEX]

[TEX]8cos^62x-12cos^42x+5cos^22x-1=0[/TEX]

 

[jerusalem]

[TEX]4.sin^{3}2x.cos6x+cos^{3}2x.sin6x=\frac{3}{8}[/TEX]

\Leftrightarrow[TEX](1-cos^2x)sin2x.cos6x+(1-sin^2x)cos2x.sin6x=\frac{3}{8}[/TEX]
\Leftrightarrow[TEX]\frac{1}{2}(1-cos^2x)(sin8x-sin4x)+\frac{1}{2}(1-sin^2x)(sin8x+sin4x)=\frac{3}{8}[/TEX]
\Leftrightarrow[TEX]\frac{1-cos4x}{2}[2sin4x(cos4x-\frac{1}{2})]+\frac{1+cos4x}{2}[2sin4x(cos4x+\frac{1}{2})]=\frac{3}{4}[/TEX]
\Leftrightarrow[TEX]sin4x[(1-cos4x)(cos4x-\frac{1}{2})+(1+cos4x)(cos4x+\frac{1}{2})=\frac{3}{4}[/TEX]
\Leftrightarrow[TEX]\frac{sin8x}{2}+2sin8x=\frac{3}{4}[/TEX]

 

[huutrang93]

1)
[TEX]sin3x+sin2x=5sinx[/TEX]

[TEX]3sinx-4sin^3x+sin2x=5sinx[/TEX]

[TEX]4sin^3x+2sinx-2sinxcosx=0[/TEX]

[TEX]2sin^3x+sinx(1-cosx)=0[/TEX]

[TEX]2sinx(1-cos^2x)+sinx(1-cosx)=0[/TEX]

[TEX](1-cosx)[2sinx(1+cosx)+sinx]=0[/TEX]

[TEX]cosx=1[/TEX]

[TEX]2sinx(1+cosx)+sinx=0[/TEX]

[TEX]3sinx+2sinxcosx=0[/TEX]

[TEX]sinx=0[/TEX]

[TEX]2cosx+3=0(L)[/TEX]

giải nghiệm cho khoảng [TEX][0;14][/TEX] là xong

cách 2

[TEX]<=> cosx+cos3x-2cos5x=0[/TEX]

[TEX]<=>2cos2xcosx-2cosxcos4x+2sinxsin4x=0[/TEX]

[TEX]<=>2cos2xcosx-2cosxcos4x+4sinxsin2xcos2x=0[/TEX]

[TEX]<=>cos2xcosx-cosxcos4x+4sin^2xcosxcos2x=0[/TEX]

[TEX]<=>cosx(cos2x-cos4x+4sin^2xcos2x)=0[/TEX]

[TEX]<=>cosx=0[/TEX]

hoặc
[TEX] cos2x-cos4x+4sin^2xcos2x=0[/TEX]

[TEX]<=> cos2x-(2cos^22x-1)+2(1-cos2x)cos2x=0[/TEX]

giải pt bậc 2=> x