f (x)=(sin^4x+cos^4x+sin^2cos^2)/(2-sin2x)
={[(sinx)^2+(cosx)^2]^2-(sinxcosx)^2}/(2-2sinxcosx)
=(1-(sinxcosx)^2)/(2-2sinxcosx)
=(1+sinxcosx)(1-sinxcosx)/2(1-sinxcosx)
=1/2+1/4sin2x
最小正周期T=π
最大值1/2+1/4=0.75
最小值1/2-1/4=0.25
單調(diào)遞增區(qū)間
(-π/4+kπ,π/4+kπ) k為任意整數(shù)