Question

函数y=√2sinx+1的定义域是

为什么是－π/6+2kπ≤x≤7π/6+2kπ而不是－π/6+2kπ≤x≤5π/6+2kπ

Answer 1

2sinx + 1≥0
2sinx≥-1
∴sinx≥-1/2
则-π/6≤x≤7π/6
即：-π/6+2kπ≤x≤7π/6+2kπ