已知：函数y=Acosx+B,(A>0)的最大值是1,最小值是-3,试确定f(x)=Bsin(ax+π/3)的单调增区间

A>0,
所以当cosx=1时,函数值最大,即A+B=1
当cosx=-1时,函数值最小,即-A+B=-3
解得,A=2,B=-1
则f(x)=Bsin(ax+π/3)=-sin(2x+π/3)
[求f(x)=-sin(2x+π/3)的单调增区间,就是求f(x)=sin(2x+π/3)的单调减区间]
令2x+π/3大于等于π/2+2Kπ,小于等于3π/2+2Kπ
解得X属于[π/12+Kπ,2π/3+Kπ](K属于整数)
则f(x)的单调增区间是[π/12+Kπ,2π/3+Kπ](K属于整数)

由题可知A+B=1 B-A=-3
进而知A=2 B=-1
f(x)=-sin(2X+π/3)
从而可得f(x)的增区间为π/12+2Kπ到7π/12+2Kπ