f(x)=m*n=2sin²x-sinx+cos2x-cosx+1
=1-cos2x-sinx+cos2x-cosx+1
=2-√2sin(x+π/4),
f(x)的最小正周期T=2π,
由-π/2+2kπ≤x+π/4≤π/2+2kπ,即-3π/4+2kπ≤x≤π/4+2kπ (k∈Z),
得sin(x+π/4)单调递增,从而f(x)单调递减,
故f(x)的单减区间是[-3π/4+2kπ,π/4+2kπ] (k∈Z),
当m*n-√2/2,
所以-π/4+2kπ