函数y=根号(cos2x)+lg(1-cosx)定义域为.大括号cos2x≥0解出-π/4+kπ<x<π/4+kπ1-cosx>0解出x≠2kπ讨论:当k=2n时,2nπ-π/4≤x≤2nπ+π/4当k=2n+1时,2nπ+3π/4≤x≤2nπ+5/4π然后怎么把解并起来啊?

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函数y=根号(cos2x)+lg(1-cosx)定义域为.大括号cos2x≥0解出-π/4+kπ<x<π/4+kπ1-cosx>0解出x≠2kπ讨论:当k=2n时,2nπ-π/4≤x≤2nπ+π/4当

函数y=根号(cos2x)+lg(1-cosx)定义域为.大括号cos2x≥0解出-π/4+kπ<x<π/4+kπ1-cosx>0解出x≠2kπ讨论:当k=2n时,2nπ-π/4≤x≤2nπ+π/4当k=2n+1时,2nπ+3π/4≤x≤2nπ+5/4π然后怎么把解并起来啊?
函数y=根号(cos2x)+lg(1-cosx)定义域为.
大括号cos2x≥0解出-π/4+kπ<x<π/4+kπ
1-cosx>0解出x≠2kπ
讨论:当k=2n时,2nπ-π/4≤x≤2nπ+π/4
当k=2n+1时,2nπ+3π/4≤x≤2nπ+5/4π
然后怎么把解并起来啊?

函数y=根号(cos2x)+lg(1-cosx)定义域为.大括号cos2x≥0解出-π/4+kπ<x<π/4+kπ1-cosx>0解出x≠2kπ讨论:当k=2n时,2nπ-π/4≤x≤2nπ+π/4当k=2n+1时,2nπ+3π/4≤x≤2nπ+5/4π然后怎么把解并起来啊?
cos2x≥0,1-cosx>0
kπ-π/4<x<kπ+π/4,x≠2kπ
∴(2k-1)π-π/4<x