Question

已知函数f(x)=sin(2x+π/6)+sin(2x-π/6)+2cos^2x +a，当x∈【-π/4，π/4】时，f（x）的最小值为-3，求a

Answer 1

f(x)=√3sin2x+cos2x+1+a
=2sin(2x+π/6)+1+a
该函数在区间【-π/3,π/6】上递增，
所以，在【-π/4，π/4】中，当X=-π/4时，f(x)有最小值：
f(x)min=2sin[2(-π/4)+π/6]
=-2cosπ/6+a+1
=-√3+a+1
=-3
所以a=-4+√3

Answer 2

就是和差化积 积化和差的应用；在x∈【-π/4，π/4】时，f(x)=sin(2x+π/6)+sin(2x-π/6)+2cos^2x +a=√3sina(2x）+cos(2x）+1+a+=2sina(2x+π/3)+1+a≧-2+1+a=-3
a=-2

Answer 3

f(x)=2sin(2x+30。)+a +1在［-45。，45。］上最小值为3，所以f(x=-45 。)=3，所以a=2+根号3