a) y=$4cos^2x-4cosx+3=(2cosx-1)^2+2 \geq 2$
=> miny=2 khi cosx=0,5 -> giải tiếp ra cosx
mặt khác $-1 \leq cosx \leq 1$
=> $cos^2x \leq 1$
$-cosx \leq 1$
=> y $\leq 4+4+3=11$ dấu = xảy ra khi cosx=-1 ...
b)y=$sin^2x+2cosx+1=-cos^2x+2cosx+2=3-(cosx-1)^2 \leq 3$
=> maxy=3 khi x=1
miny khi $(cosx-1)^2$ max <=> $|cosx-1|$ max
mà $-1 \leq x \leq 1$ => |cosx-1| max = 2 khi cosx=-1 => min y=-1 khi cosx=-1