高一数学题，求详细过程
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原式=2cosx(sinxcos兀/3-cosxsin兀/3)+√3sin^2 x +sinxcosx
=sinxcosx -√返段3cos^2 x +√3sin^2 x +sinxcosx
=2sinxcosx-√3(cos^2 x-sin^2 x)
=sin2x-√3cos2x
=2(sin2xcos兀/3-cos2xsin兀/3)
=2sin(2x-兀/3)
T最小弊世戚=2兀/2=兀
2
2k兀-兀/2<=2x-兀/3<=2k兀+兀/2
2k兀-兀/6<=2x<=2k兀+5兀/6
k兀-兀/12<=x<=k兀+5兀租陵/12
=sinxcosx -√返段3cos^2 x +√3sin^2 x +sinxcosx
=2sinxcosx-√3(cos^2 x-sin^2 x)
=sin2x-√3cos2x
=2(sin2xcos兀/3-cos2xsin兀/3)
=2sin(2x-兀/3)
T最小弊世戚=2兀/2=兀
2
2k兀-兀/2<=2x-兀/3<=2k兀+兀/2
2k兀-兀/6<=2x<=2k兀+5兀/6
k兀-兀/12<=x<=k兀+5兀租陵/12
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