求函数f(x)=cos2x+cos(2x-π/3)的最小值及取到最小值时x的值希望能够有详细的过程,不光是一个答案就够了.

解；f(x)=cos2x+cos(2x-π/3)＝cos2x+1/2cos2x+√3/2sin2x
=3/2cos2x+√3/2sin2x=√3(√3/2cos2x+1/2sin2x)
＝√3sin(2x+π/3)
故当2x+π/3＝2kπ-π/2,即 x=kπ-5π/12,k∈Z时，函数有
最小值－√3

f(x)=cos2x+cos(2x-π/3)
=cos2x+cos2xcosπ/3+sin2xsinπ/3
=3cos2x/2+√3sin2x/2
=√3(√3cos2x/2+sin2x/2)
=√3sin(2x+π/3)
最大值为√3,2x+π/3=2kπ+π/2,x=kπ+π/12
最小值为-√3,2x+π/3=2kπ+3π/2,x=kπ+7π/12

f(x)=cos2x+cos(2x-π/3)=cos2x+cos2xcosπ/3+sin2xsinπ/3=3/2cos2x+√3/2sin2x=√3(1/2sin2x+√3/2cos2x)=√3sin(2x+π/3)
因为-1

f(x)=cos2x+cos(2x-π/3)=cos2x+1/2 cos2x +根号3/2 sin2x
=根号3 sin(2x+π/3)
最小值是 -根号3
2x+π/3=π+2kπ，k属于z
x=π/3x+kπ,k属于z