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[TEX]1/ cosx + cos3x = tan\frac{x}{4} + cot\frac{x}{4}[/TEX]
[TEX]PT \Leftrightarrow 2cos2x.cosx=\frac{1}{sin(\frac{x}{4}).cos(\frac{x}{4})} \Leftrightarrow 2cos2x.cosx=\frac{2}{sin(\frac{x}{2}) \Leftrightarrow cos2x.cosx.sin(\frac{x}{2})=1 \Leftrightarrow \left[\begin{|cos2x|=1}\\{|cosx| = 1}\\{|sin(\frac{x}{2})|=1}[/TEX]
[TEX]2/ 8cot^2 x + 2 tan ^8 x = 10[/TEX]
Áp dụng BĐT Cauchy ta có:
[TEX]8cot^2 x + 2 tan ^8 x=cot^2x+cot^2x+cot^2x+cot^2x+cot^2x+cot^2x+cot^2x+cot^2x+tan^8x+tan^8x \geq 10[/TEX]
[TEX]PT \Leftrightarrow cot^2x=ta^2x \Leftrightarrow ...[/TEX]
[TEX]3/ 3sinx + 4cosx = 5 + ( 4 tanx - 3 )^2[/TEX]
[TEX](3sinx + 4cosx)^2 \leq (3^2+4^2)(sin^2x+cos^2x)=25 \Rightarrow 3sinx+4cosx \leq 5[/TEX] Đẳng thức xảy ra \Leftrightarrow 4sinx=3cosx
[TEX]5 + ( 4 tanx - 3 )^2 \geq 5[/TEX] đẳng thức xảy ra \Leftrightarrow 4tanx-3=0
[TEX]PT \Leftrightarrow \left{\begin{4sinx=3cosx}\\{4tanx-3=0}[/TEX]
[TEX]4/ cos^5 x - sin^5 x = sinx - cosx [/TEX]
[TEX]PT \Leftrightarrow cosx(1-cos^4x)=sinx(1-sin^4x) \Leftrightarrow cosx(1-cos^2x)(1+cos^2x)=sinx(1-sin^2x)(1+sin^2x) \Leftrightarrow cosx.sin^2x(1+cos^2x)=sinx.cos^2x(1+sin^2x) \Leftrightarrow sinx.cosx[sinx+sinx.cos^2x-cosx-cosx.sin^2x]=0 \Leftrightarrow sin2x.(sinx-cosx)(1-sinx-cosx)=0 \Leftrightarrow ....[/TEX]
[TEX]5/ cos^6 x + sin^6 x = 2 (cos^8 x + sin ^8 x)[/TEX]
[TEX]PT \Leftrightarrow cos^6x(2cos^2x-1)=sin^6x(1-2sin^2x)=0 \Leftrightarrow cos^6x.cos2x=sin^6x.cos2x \Leftrightarrow cos2x(sin^6x-cos^6x)=0 \Leftrightarrow ....[/TEX]
[TEX]6/ cos^6 x + sin^6 x = cos4x[/TEX]
[TEX]cos^6x+sin^6x=1-\frac{3}{4}sin^22x[/TEX]
[TEX]cos4x=1-2sin^22x[/TEX]
[TEX]7/ \frac{sin^4 x + cos^4 x }{sin2x} = \frac{1}{2} ( tanx + cotx)[/TEX]
[TEX]sin^4x+cos^4x=1-\frac{1}{2}sin^22x[/TEX]
[TEX]tanx+cotx=\frac{2}{sin2x}[/TEX]
[TEX]8/ ( sinx + 3 ) sin ^4 \frac{x}{2} - ( sinx + 3 )sin^2 \frac{x}{2} +1 = 0[/TEX]
[TEX]sin^4(\frac{x}{2})=\frac{1}{4}(1-cosx)^2[/TEX]
[TEX]sin^2(\frac{x}{2})=\frac{1-cosx}{2}[/TEX]