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f(x)=√3cos²;x+sinxcosx =√3/2*(cos2x+1)+1/2(sin2x) =1/2sin2x+√3/2cos2x+√3/2 =sin(2x+π/3)+√3/2 當2x+π/3=2kπ+π/2,k∈Z,取最大值1+√3/2; 當2x+π/3=2kπ-π/2,k∈Z,取最大值√3/2-1.
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