f'(x)=cosx+rad3sinx.
1)f'(x)=0
cosx+rad3sinx=0 :cosx
1+3tgx=0
tgx=-radical din 3/3
x=-pi/6 +mpi, m apartine lui Z.
pi/2<= -pi/6+mpi<=pi
2pi/3<=mpi<=7pi/6
2/3<=m<=7/6
m=1, x=-pi/6+pi=5pi/6
2) f(5pi/6)= sin(pi-pi/6) -rad3 cos(pi-pi/6)+1= 1/2+3/2+1=3.
f(pi/2)=1-0+1=2
f(pi)=rad3+1
Raspuns: m=min=2, M=max=3.
