已知函数f(x)=cos(π/3-x). (1)求函数f(x)的单调递增区间;(2)若f(x)>=更号3/2,求x的取值范围.
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人气:290 ℃ 时间:2019-08-18 21:57:49
解答
(1)f(x)=cos(π/3-x)则,f(x)=cos(x-π/3) 递增区间:kπ-π/2≤x-π/3≤kπ kπ-π/6≤x≤kπ+π/3 (k是整数)(2)f(x)=cos(x-π/3) ≥√3/2=cos(kπ+π/3)kπ-π/3≤x-π/3≤kπ+π/3kπ≤x-π/3≤kπ+2π/3 (k是...
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