9-x²>=0
-3<=x<=3
√3 -(√3 -1)tanx-tan²x>0
tan²x+(√3-1)tanx-√3<0
(tanx+√3)(tanx-1)<0
-√3kπ-π/3k=-1
π≈3.14
所以-4.19-3<=x<=3
所以-3<=x<-3π/4
k=0
-1.05在[-3,3]范围内
同理验证k=1
所以
[-3,-3π/4)∪(-π/3,π/4)∪(2π/3,3]