1)sinx≤根号2/2, 解得{x│2kπ-5π/4≤x≤2kπ+π/4,k∈Z}
2)cosx≥-1/2 解得{x│2kπ-2π/3≤x≤2kπ+2π/3, k∈Z}
sinx≥根号3/2 解得{x│2kπ+π/3≤x≤2kπ+2π/3, k∈Z}
则{x│2kπ-2π/3≤x≤2kπ+2π/3, k∈Z}∪{x│2kπ+π/3≤x≤2kπ+2π/3, k∈Z}
={x│2kπ+π/3≤x≤2kπ+2π/3, k∈Z}
(1){x|-5π/4+2kπ≤x≤π/4+2kπ,k∈Z};
(2){x|π/3+2kπ≤x≤2π/3+2kπ,k∈Z}。