y=x+|sin2x|确定函数单调区间方法1y=x+|sin2x|等价于 y=x+sin2x kπ≤x≤kπ+π/2 y=x-sin2x kπ+π/2≤x≤(k+1)π对该曲线方程求导,得 y'=1+2cos2x kπ≤x≤kπ+π/2 y'=1-2cos2x kπ+π/2≤x≤(k+1)π当y'>0时,该函数单调

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