Question

已知函数f(x)=sin(2x+π/6)+sin(2x-π/6)+cos2x+a(a∈R,a为常数）

1、求f(x)的最小正周期
2、求单调递增区间
3、若x∈[0,π/2]，f(x)的最小值是-2，求a的值

Answer 1

1.
f(x)=sin(2x+π/6)+sin(2x-π/6)+cos2x+a
=√3/2sin2x+1/2cos2x+√3/2sin2x-1/2cos2x+cos2x+a
=√3sin2x+cos2x+a
=2sin(2x+π/6)+a
所以f(x)的最小正周期2π/2=π
2.
sinx的单调递增区间[2kπ-π/2，2kπ+π/2]
所以2kπ-π/2<=2x+π/6<=2kπ+π/2
解得kπ-π/3<=x<=kπ+π/6
单调递增区间[kπ-π/3,kπ+π/6] k是整数
3.
x∈[0,π/2] 2x+π/6∈[π/6,7π/6]
sin(2x+π/6)的最小值是sin(7π/6)=-1/2
所以2*(-1/2)+a=-2 a=-1

Answer 2

f(x)=sin(2x+π/6)+sin(2x-π/6)+cos2x+a
=2sin2xcosπ/6+cos2x+a
=2sin(2x+π/6)+a
1)T=π
2)sinx单调递增区间为[2kπ-1/2π,2kπ+1/2π]