Question

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

Answer 1

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