$y = \sin^6x + \cos^6x$
$= (\sin^2x + \cos^2x)(\sin^4x - \sin^2x\cos^2x + \cos^4x)$
$= (\sin^2x + \cos^2x)^2 - 3\sin^2x\cos^2x$
$= 1 - \dfrac{3}{4}\sin^22x$
Ta có: $0 \leq \sin^22x \leq 1$
$\Leftrightarrow -\dfrac{3}{4}\leq -\dfrac{3}{4}\sin^22x \leq 0$
$\Leftrightarrow \dfrac{1}{4} \leq 1 - \dfrac{3}{4}\sin^22x \leq 1$
Hay $\dfrac{1}{4} \leq y \leq 1$
Vậy $\min y = \dfrac{1}{4} \Leftrightarrow \sin^22x = 1 \Leftrightarrow x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}$
$\max y = 1 \Leftrightarrow \sin2x = 0 \Leftrightarrow x = k\dfrac{\pi}{2} \quad (k \in \Bbb Z)$
