急求一道数学题的答案，谢谢

f(x)=2sin²(π/4-x)-(√3)cos2x=1-cos(π/2-2x)-(√3)cos2x=1-sin2x-(√3)cos2x
=1-2[(1/2)sin2x-(√3/2)cos2x]=1-2[sin2xcos(π/3)+cos2xsin(π/3)]=1-2sin(2x+π/3);
(1). f(x)的最小正周期T=2π/2=π；
单调减区间：由2kπ-π/2≦2x+π/3≦2kπ+π/2，,即2kπ-5π/6≦2x≦2kπ+π/6，
得单减区间为：kπ-5π/12≦x≦kπ+π/12；(K∈Z)
单调增区间：由2kπ+π/2≦2x+π/3≦2kπ+3π/2，即2kπ+π/6≦2x≦2kπ+7π/6；
得单增区间：kπ+π/12≦x≦kπ+7π/12; (k∈Z)
(2). 当x=π/2+2kπ时f(x)获得最大值=1-2sin(4kπ+π+π/3)=1+2sin(π/3)=1+2×(√3/2)=1+√3
当x=π/12+2kπ时f(x)获得最小值=1-2sin(4kπ+π/6+π/3)=1-2sin(π/2)=1-2=-1.

图看不清

我见过你这样的好几次了，都是马赛克的高清大图，我是好心路人甲，删掉发原图

18还是1918. (1)∠A∠D (2)∠ABD ∠CBD ∠ ABC (3)∠A ∠B ∠ABC ∠ABD ∠CBD ∠BDC ∠BDA19.3个；6个；10个；66个