mình nghĩ thế này :
[TEX]tg^2x+cot^2x \geq 2\sqrt{(tgxcotx)^2}=2[/TEX]
[TEX]2sin^5(x-\frac{\pi}{4}) \leq 2[/TEX]
[TEX]pt \Leftrightarrow[/TEX]
[TEX]\left| {\begin{tg^2 x + \cot ^2 x = 2}\\{2\sin ^5 (x - \frac{\pi }{4}) = 2}[/TEX]
mình có cách #
[TEX]tg^2x+cot^2x=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}=\frac{1-cos2x}{1+cos2x}+\frac{1+cos2x}{1-cos2x}=\frac{2(1+cos^22x)}{sin^22x}=2(1+cot^22x)[/TEX]
pt\Leftrightarrow[TEX](1+cot^22x)=sin^5(x-\frac{\pi}{4})[/TEX]
ta có
[TEX](1+cot^22x)\geq1[/TEX]
[TEX]sin^5(x-\frac{\pi}{4})\leq sin(x-\frac{\pi}{4})\leq1[/TEX]
vậy dấu =
[TEX]\left| {\begin{cot^22x= 0}\\{\sin ^5 (x - \frac{\pi }{4}) = 1}[/TEX]
\Leftrightarrow[TEX]x=\frac{\pi}{4}+\frac{kpi}{2}[/TEX]