f(x)=2(cos^2x+2·√3/2cosx+3/4)+a-3/2=2(cosx+√3/2)^2+a-3/2(cosx+√3/2)^2≥0则当(cosx+√3/2)^2=0 时f(x)取得最小值,f(x)min=a-3/2当x=0时,(cosx+√3/2)^2取最大值,f(x)取得最大值,f(x)max=f(0)=a+2+2√3,由题...当(cosx+√3/2)^2=0 时f(x)取得最小值，f(x)min=a-3/2当x=0时，(cosx+√3/2)^2取最大值，f(x)取得最大值，f(x)max=f(0)=a+2+2√3注意定义域，谢谢我的错(cosx+√3/2)^2=0，x想根本不能满足那只能在x=π/3取最小值了f(x)min=f(π/3)=1/2+√3+af(x)max+f(x)min=a+2+2√3+1/2+√3+a=3a=1/4-3√3/2