一道高中三角函数题，求大神详细过程，谢谢！
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(1)
f(x)=(√3/2)sin2x+cos²x-½
=(√3/2)sin2x+½(2cos²x-1)
=(√3/2)sin2x+½cos2x
=sin2xcosπ/6+cos2xsinπ/6
=sin(2x+ π/6)
最小正周期T=2π/2=π
(2)
f(x)=2sinxcosx+2√3cos²x-√3
=sin2x+√3(2cos²x-1)
=sin2x+√3cos2x
=2[½sin2x+(√3/2)cos2x]
=2(sin2xcosπ/3+cos2xsinπ/3)
=2sin(2x+ π/3)
最小正周期T=2π/2=π
f(x)=(√3/2)sin2x+cos²x-½
=(√3/2)sin2x+½(2cos²x-1)
=(√3/2)sin2x+½cos2x
=sin2xcosπ/6+cos2xsinπ/6
=sin(2x+ π/6)
最小正周期T=2π/2=π
(2)
f(x)=2sinxcosx+2√3cos²x-√3
=sin2x+√3(2cos²x-1)
=sin2x+√3cos2x
=2[½sin2x+(√3/2)cos2x]
=2(sin2xcosπ/3+cos2xsinπ/3)
=2sin(2x+ π/3)
最小正周期T=2π/2=π
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