$y=\sin^4x+\cos^4x$
$=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x$
$=1-\dfrac{1}{2}(\sin2x)^2$
$=\dfrac{-1}{2}\sin^22x+1$
Ta có: $0\le \sin^22x\le 1$
$\to \dfrac{-1}{2}\le -\dfrac{1}{2}\sin^22x\le 0$
$\to \dfrac{1}{2}\le y\le 1$
Vậy $\min y=\dfrac{1}{2}; \max y=1$
$y=\cos x-\sqrt3\sin x$
$=2\cos\left(x+\dfrac{\pi}{3}\right)$
Ta có $\cos\left(x+\dfrac{\pi}{3}\right)\in [-1;1]$
$\to y\in [-2;2]$
Vậy $\min y=-2;\max y=2$
