已知函数f(x)=sin(2x+π/6)-cos2x

题目

已知函数f(x)=sin(2x+π/6)-cos2x
1 求f(x)的最小正周期和单调递增区间
2 求f(x)在【0,π/2】上的最小值和最大值及相应的x值
3 若函数f(x)满足方程f(x)=a(0＜a＜1),求在[0,2π]内的所有实数根之和.

答案

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