$sin^6x+cos^6x=(sin^2x+cos^2x)^3-3sin^2x.cos^2x(sin^2x+cos^2x)$=$1-3sin^2x.cos^2x=1-\frac{3}{4}sin^22x=1-\frac{3}{4}.\frac{(1-cos4x)}{2}$$=1-\frac{3}{8}(1-cos4x)=1-\frac{3}{8}+\frac{3}{8}cos4x=\frac{5+3cos4x}{8}$
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y=sin^6x+cos^6x=(sin^2x+cos^2x)^3-3sin^2x.cos^2x(sin^2x+cos^2x)$=$1-3sin^2x.cos^2x=1-\frac{3}{4}sin^22x=1-\frac{3}{4}.\frac{(1-cos4x)}{2}$$=1-\frac{3}{8}(1-cos4x)=1-\frac{3}{8}+\frac{3}{8}cos4x=\frac{5+3cos4x}{8}$
+)Vì $cos4x\geq -1\Rightarrow y\geq \frac{5+3.(-1)}{8}=1/4$ dấu "=" xảy ra $\Leftrightarrow cos4x=-1\Leftrightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}$($k\in Z)$+)$cos4x\leq 1\Rightarrow y\leq \frac{5+3}{8}=1$dấu "=" xảy ra $\Leftrightarrow cos4x=1\Leftrightarrow x=\frac{k\pi}{2}(k\in Z)$Vậy...