题目

已知函数f（x）=sin的平方（2x+4分之π）+（根号3）cos的平方2x 求f（x）的最小正周期和单调
递减区间（2）若x∈【-6分之π,6分之π】,求f（x）的最值及最值时相应的x的值

提问时间：2020-11-13

答案

f（x）= sin^2(2x+π/4) + √3 cos^2(2x)
= {sin2xcosπ/4+cos2xsinπ/4}^2 + √3 cos^2(2x)
= 1/2(sin2x+cos2x)^2 + √3 cos^2(2x)
= 1/2(1+2sin2xcos2x) + √3 * (cos4x+1)/2
= 1/2 + 1/2sin4x +√3/2 cos4x + √3/2
= sin4xcosπ/3 + cos4xsinπ/3 + (1+√3)/2
= sin(4x+π/3) + (1+√3)/2
最小正周期=2π/4=π/2
当4x+π/3∈（2kπ+π/2,2kπ+3π/2),其中k∈Z时单调减
单调减区间：x∈（kπ/2+π/24,kπ/2+7π/24),其中k∈Z
若x∈【-π/6,π/6】
4x+π/3∈（-π/3,π)
4x+π/3=-π/3,x=-π/6时,函数有最小值f(x)min=sin(-π/3) + (1+√3)/2 = -√3/2+(1+√3)/2 = 1/2
4x+π/3=π/2,x=π/24时,函数有最大值f(x)max=sin(π/2) + (1+√3)/2 = 1+(1+√3)/2 =(3+√3)/2