∵y＝lg(2sinx−1)+1−2cosx∴2sinx-1＞0    ①1-2cosx≥0      ②由①得，sinx＞12由②得cosx≤12，∴π3+2kπ≤x＜5π6+2kπ，k∈z故答案为：[π3+2kπ，5π6+2kπ)...