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数学
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函数f(X)=cos^x-sin^2的单调递增区间
人气:127 ℃ 时间:2020-02-03 05:44:20
解答
f(X)=cos²x-sin²x=cos2x
令 -π+2kπ≤2x≤2kπ
得 -π/2+kπ≤x≤kπ
所以 单调递增区间为[-π/2+kπ,kπ] ,k∈Z
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