f(x)=3sin(x+π/3)+√3cos[π/2-(π/6-x)]=3sin(x+π/3)+√3cos(x+π/3)=2√3[√3/2*sin(x+π/3)+1/2*cos(x+π/3)]=2√3[sin(x+π/3)cosπ/6+cos(x+π/3)sinπ/6]=2√3sin(x+π/3+π/6)sin(x+π/3+π/6)最大=1所以f(x)最大=2√3
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