求函数y=√（2cosx-1)(1-cosx)+lg(36-x²）的定义域

真数大于0，36-x²>0
-6<x<6

根号下大于等于0
(2cosx-1)(1-cosx)>=0
cosx<=1,所以1-cosx>=0
所以2cosx-1>=0
cosx>=1/2=cos(2kπ±π/3)
所以2kπ-π/3<=x<=2kπ+π/3

k=-1，-7π/3<-6<-5π/3
k=0，-6<-π/3<π/3<6
k=1,5π/3<6<7π/3

综上
定义域是(-6,-5π/3]∪[-π/3,π/3]∪[5π/3,6)