f(x)=5/2*sin2x-5√3*(1+cos2x)/2+5√3/2
=5/2*sin2x-5√3/2*cos2x
=5(sin2x*1/2-cos2x*√3/2)
=5(sin2xcosπ/3-cos2xsinπ/3)
=5sin(2x-π/3)
sinx增则2kπ-π/2
所以单调增加则
2kπ-π/2<2x+π/6<2kπ+π/2
2kπ-2π/3<2x<2kπ+π/3
kπ-π/3
同理，减区间(kπ+π/6,kπ+2π/3)
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