$y=\sin^4\dfrac{x}{2}+\cos^4\dfrac{x}{2}-\sin^2\dfrac{x}{2}\cos^2\dfrac{x}{2}$

$=\Big( \sin^2\dfrac{x}{2}+\cos^2\dfrac{x}{2}\Big)^2-3\sin^2\dfrac{x}{2}\cos^2\dfrac{x}{2}$

$=1-\dfrac{3}{4}\sin^2x$

Có: $0\le \sin^2x\le 1$

$\to \dfrac{-3}{4}\le -\dfrac{3}{4}\sin^2x\le 0$

$\to \dfrac{1}{4}\le y\le 1$

Vậy:

$\max y=1\to \sin^2x=0\to x=k\pi$

$\min y=\dfrac{1}{4}\to \sin^2x=1\to \cos x=0\to x=\dfrac{\pi}{2}+k\pi$

`~rai~`

\(3)y=sin^6\dfrac{x}{2}+cos^6\dfrac{x}{2}\\\quad=\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}\right)\left(sin^4\dfrac{x}{2}-sin^2\dfrac{x}{2}cos^2\dfrac{x}{2}+cos^4\dfrac{x}{2}\right)\\\quad=sin^4\dfrac{x}{2}-sin^2\dfrac{x}{2}cos^2\dfrac{x}{2}+cos^4\dfrac{x}{2}\\\quad=\left(sin^4\dfrac{x}{2}+2sin^2\dfrac{x}{2}cos^2\dfrac{x}{2}+cos^4\dfrac{x}{2}\right)-3sin^2\dfrac{x}{2}cos^2\dfrac{x}{2}\\\quad=\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}\right)-3.\dfrac{1}{4}.4sin^2\dfrac{x}{2}cos^2\dfrac{x}{2}\\\quad=1-\dfrac{3}{4}sin^2x.\\\text{Ta có:}0\le sin^2x\le 1\\\Leftrightarrow -\dfrac{3}{4}\le -\dfrac{3}{4}sin^2x\le 0\\\Leftrightarrow \dfrac{1}{4}\le 1-\dfrac{3}{4}sin^2x\le 1\\\Leftrightarrow \dfrac{1}{4}\le y\le 1.\\\Rightarrow \begin{cases}Min_{y}=\dfrac{1}{4}\\Max_{y}=1.\end{cases}\\+)Min_{y}=\dfrac{1}{4}\Leftrightarrow sin^2x=0\Leftrightarrow sinx=0\Leftrightarrow x=k\pi.(k\in\mathbb{Z}).\\+)Max_{y}=1\Leftrightarrow sin^2x=1\Leftrightarrow cos^2x=0\Leftrightarrow cosx=0\Leftrightarrow x=\dfrac{x}{2}+k\pi.(k\in\mathbb{Z}).\)