解f(x)=2sin^2x-sin(π+2x),=2sin^2x+sin(2x)=2（1-cos2x）/2+sin(2x)=1-cos2x+sin(2x)=sin(2x)-cos2x+1=√2sin（2x-π/4)+1即周期T=2π/2=π由-1≤sin（2x-π/4)≤1即-√2≤√2sin（2x-π/4)≤√2即-√2+1≤√2sin...