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求下列函数的单调区间(1)y=sin(3x-π/3)(2)y=-cos(2x-π/3)
人气:285 ℃ 时间:2020-01-30 02:12:13
解答
(1)由图像可得,当-π/2+2kπ≤3x-π/3≤π/2+2kπ,即-π/18+2kπ/3≤x≤5π/18+2kπ/3时函数单调递增;
同理,当π/2+2kπ≤3x-π/3≤3π/2+2kπ,即5π/18+2kπ/3≤x≤11π/18+2kπ/3时函数单调递减.其中k∈Z.
(2)由图像可得,-π/2+2kπ≤2x-π/3≤2kπ,即-π/12+kπ≤x≤π/6+kπ时函数单调递减;
同理,当2kπ≤2x-π/3≤π/2+2kπ,即π/6+kπ≤x≤5π/12+kπ时函数单调递增.其中K∈Z.
这里把函数看成y=sinA和y=-cosB,这样问题就简单了.
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