f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a,求单调递增区间,若在[-π/2,π/2]上最大值和最小值和为根号3求a值
问题描述：
f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a,求单调递增区间,若在[-π/2,π/2]上最大值和最小值和为根号3求a值
答
f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a,
=√3sinx+cosx+a
=2sin(x+π/6)+a
单调递增区间：2kπ-π/2≤x+π/6≤2kπ+π/2
得：2kπ-2π/3≤x≤2kπ+π/3
若在[-π/2,π/2]上最大值和最小值和为根号3
当x=-π/2有最小值：y=-√3+a
当x=π/3有最大值：y=2+a
于是有：-√3+a+2+a=√3
解得：a=√3+1