g,
$y=3(1-2\sin^2x)-\sin^2x$
$=-7\sin^2x+3$
Ta có $\sin^2x\in[0;1]$
$\to -7\sin^2x\in[-7;0]$
$\to y\in[-4;3]$
Vậy $\min y=-4; \max y=3$
j,
$y=\cos^2x+2\sin x+2$
$=1-\sin^2x+2\sin x+2$
$=-\sin^2x+2\sin x+3$
Đặt $t=\sin x$ ($t\in [-1;1]$)
$\to y=-t^2+2t+3$
Đỉnh parabol: $\left( \dfrac{-2}{-2}; \dfrac{-4.3-2^2}{-4}\right)=(1;4)$
$\to \max y=y(1)=-1+2+3=4; \min y=y(-1)=-1-2+3=0$ (do parabol có $a<0$)
l,
$y=\cos2x-\sqrt3\sin2x+2$
$=2\cos\left(2x+\dfrac{\pi}{3}\right)+2$
Ta có $\cos\left(2x+\dfrac{\pi}{3}\right)\in [-1;1]$
$\to y\in[0;4]$
Vậy $\min y=0; \max y=4$
