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

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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
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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
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

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
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

f(x)=√3sinx/2+cosx/2+√3sinx/2-cosx/2+cosx+a
=√3sinx+cosx+a
=2sin(x+π/6)+a
所以最大值为2+a，最小值为-2+a
2+a+(-2+a)=√3
2a=√3
a=√3/2

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