已知函数f(x)=sin(2x+π⼀6)-cos2x

f(x)=sin(2x+π/6)+cos(2x)
=sin(2x)cos(π/6)+cos(2x)sin(π/6)+cos(2x)
=(√3/2)sin(2x)+(1/2)cos(2x)+cos(2x)
=√3((1/2)sin(2x)+(√3/2)cos(2x))
=√3cos(2x+π/6)
（1）f(x)的最小正周期为π,单调递增区间kπ-π/2<=X<=kπ+(5π)/12
（2）画出函数曲线，很容易就看出最大值和最小值，最小值是-√3，取值点x=（5π）/12,最大值是3/2,取值点x=3/2
（3）第三条根据化简的算，也可以画曲线图知道。
计算结果kπ+1/2(arccos(√3/3)-π/6)可能计算有勿，重算下保险吧

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
单调递增区间 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时，f(x)最大=1
2x-π/3=3π/2 x=11π/12时，f(x)最小=-1
3. a>0 sin(2x-π/3)>0
在[0,2π]内
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
故所有实数根之和=π/6+(1/2)arcsina+2π/3-(1/2)arcsina
=5π/6