求函数y=根号下（-2cos2x+3cosx-1）+lg(36-x ^2)的定义域

题目

求函数y=根号下（-2cos2x+3cosx-1）+lg(36-x ^2)的定义域

答案

-2cos2x+3cosx-1>=0
cos2x-3cosx+1<=0
2cos²x-1-3cosx+1<=0
cosx(2cosx-3)<=0
cosx<=1则2cosx-3<0
所以cosx>=0
2kπ-π/2<=x<=2kπ+π/2
36-x²>0
-6取交集
是(-6,-3π/2]∪[-π/2,π/2]∪[3π/2,6)