不等式2sin+根号2≥0的解集是
1个回答

sinx ≥ - (√2)/2

-√2/2=sin(-π/4)=sin(-3π/4)

=sin(2kπ-π/4)=sin(2kπ-3π/4)

即sinx≥sin(2kπ-π/4)

sinx≥sin(2kπ-3π/4)

sinx在(2kπ-π,2kπ-π/2)是减函数

在(2kπ-π/2,2kπ)是增函数

所以

x∈(2kπ-π,2kπ-π/2)时,2kπ-3π/4≤x≤2kπ-π/2

x∈(2kπ-π/2,2kπ)时,2kπ-π/2≤x≤2kπ-π/4

综上

2kπ-3π/4≤x≤2kπ-π/4