f(x)=-2sinxcosx+2(cosx)^2+1且f(x)-1=0在（0,π）内有两个零点x1,x2；f(x)-1=-2sinxcosx+2(cosx)^2=2(cosx)^2（tanx-1）=0 cosx=0或tanx-1=0但cosx=0函数不成立得出x1=π/4 x2=3π/4
x1+x2=π
f(x)=-2sinxcosx+2(cosx)^2+2-1=2-(1-2(cosx)^2)-sin2x=2-cos2x-sin2x=2-根号下2*sin(π
/4+2x)堆成点位（π/8,0）移动为（0,2）m最小值是π/8