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}。
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