f(x)=sin(2x+π/6)-cos2x
=(√3/2)sin2x-(1/2)cos2x
=sin(2x-π/3)
1.最小正周期=2π/2=π
單調(diào)遞增區(qū)間 2kπ-π/2≤2x-π/3≤2kπ+π/2
解得x∈[kπ-π/12,kπ+5π/12]
2.2x-π/3=π/2 x=5π/12時,f(x)最大=1
2x-π/3=3π/2 x=11π/12時,f(x)最小=-1
3.a>0 sin(2x-π/3)>0
在[0,2π]內(nèi)
sin(2x-π/3)=a
2x-π/3=arcsina x=π/6+(1/2)arcsina
或2x-π/3=π-arcsina x=2π/3-(1/2)arcsina
故所有實數(shù)根之和=π/6+(1/2)arcsina+2π/3-(1/2)arcsina
=5π/6