題目出錯(cuò)了
cos（π/6=x）里面是加的還是減的?
y=2sin（π/3-x）-cos（π/6+x）
=2sin(π/3-x)-sin[π/2-(π/6+x)]
=2sin(π/3-x)-sin(π/3-x)
=sin(π/3-x)
=-sin(x-π/3)
T=2π/1=2π
最大值是1,最小值是-1
求單調(diào)遞增區(qū)間
即是求sin(x-π/3)的單調(diào)遞減區(qū)間
令2kπ+π/2＜x-π/3＜2kπ+3π/2(k∈Z)
得2kπ+5π/6＜x＜2kπ+11π/6(k∈Z)
即單調(diào)遞增區(qū)間是(2kπ+5π/6,2kπ+11π/6)(k∈Z)
求對(duì)稱軸
令y=-sin(x-π/3)=±1
那么sin(x-π/3)=±1
故x-π/3=kπ+π/2(k∈Z)
所以x=kπ+5π/6(k∈Z)
如果不懂,請(qǐng)Hi我,祝學(xué)習(xí)愉快!