求函数f(x)=cosx的四次方-2cos2x的二次方+sinx的平方的值域和最小正周期

f(x)=cos^4x-2(cos2x)^2+sin^2x=cos^4x-2(2cos^2x-1)^2+1-cos^2x=cos^4x-8cos^4x+8cos^2x-cos^2x-1=-7(cos^2x-1/2)^2+3/4=-7/4* cos^2*2x+3/4所以f(x)∈〔-1,3/4〕T=π/2