（1）求f（x）=sin^4x-cos^4x的最小正周期

解
f(x)=sin^4x-cos^4x
=(sin²x+cos²x)(sin²x-cos²x)
=sin²x-cos²x
=-cos2x
∴最小正周期为：T=2π/2=π
f(x)=2cos²x+2√3sinxcosx
=√3sin2x+(2cos²x-1)+1
=√3sin2x+cos2x+1
=2(√3/2sin2x+1/2cos2x)+1
=2(sin2xcosπ/6+cos2xsinπ/6)+1
=2sin(2x+π/6)+1
∴周期为：T=2π/2=π