f(x)=5/2sin2x-5√3/2(1+cos2x)+5√3/2
=5/2sin2x-5√3/2cos2x
=5sin(2x-π/3)
所以f(x)的最小正周期为T=2π/2=π
令2x-π/3 ∈[2kπ-π/2,2kπ+π/2]
可得递增区间 [kπ-π/12,kπ+5π/6];
f(x)=5sin(2x-π/3)=5sin[2(x-π/6)]
函数图像可由y=5sin2x向右移动π/6得到.