∵√3/2-√3sin²x+sinxcosx=(√3/2)(1-2sin²x)+(1/2)sin2x=sin(π/3)cos2x+cos(π/3)sin2x=sin(2x+π/3),∴原式=sin(2x+π/3)+(√3/2-√3sin²x+sinxcosx)=2sin(2x+π/3).
化简sin(2x+π/3)-（根号3）sin²x+sinxcosx+(根号3)/2
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