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f(x)=sin(x-π/4)+3cos(x+π/4)的单调递增区间
人气:397 ℃ 时间:2020-06-18 12:34:53
解答
f(x)=sin(x-π/4)+3cos(x+π/4)
=-cos(π/2+x-π/4)+3cos(x+π/4)
=-cos(x+π/4)+3cos(x+π/4)
=2cos(x+π/4)
π+2kπ≤x+π/4≤2π+2kπ
3π/4+2kπ≤x≤7π/4+2kπ
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