Question

y=√-sinx的定义域为

Answer 1

y=√(-sinx)的定义域为
解：由-sinx≧0，得sinx≦0，故定义域为：π+2kπ≦x≦2(k+1)π，k∈Z.

Answer 2

-sinx>=0
sinx<=0
所以
定义域为:
{x|2kπ+π<=x<=2kπ+2π}