函数f(x)=cos2x+cos(x+π/3)+sin(x+π/6)+3sin^2x的最小值

cos2x+cos(x+π/3)+sin(x+π/6)+3sin^2x
=cos2x+1/2*cosx-根号3/2*sina+根号3/2*sinx+1/2*cosx+3sin^2x
=cos2x+cosx+3sin^2x
=1-2sin^2x+cosx+3sin^2x
=1+sin^2x+cosx
=-cosx^2+cosx+2
=-(cosx-1/2)^2+9/4
当cosx=-1时,
所取得值最小为0