f(x)=sin^2x+cosx+3/4

x属于[0,2/3π],f(x)=sin^2x+cosx+3/4=1-cos²x+cosx+3/4=-(cos²x-cosx+1/4)+2=-(cosx-1/2)²+2当cosx=1/2时,f(x)最大值f=2当cosx=-1时,f(x)最小值=-1/4f(x)=sin^2x+cosx+3/4,x属于[0,2/3π],值域为[-1...