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f'(x)=0
√3cosx-sinx=0
2*cos(π/6)*cosx-sinx*sin(π/6)=0
2(cosx+π/6)=0
cos(x+π/6)=0
x+π/6=π/2+2kπ v x+π/6=-π/2+2kπ
x=-2π/3+kπ , k∈C
min=f(-2π/3)=√3*(-√3/2)-1/2=-2
max=f(π/3)=√3*√3/2+1/2=2