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函数y=sin(2x-π÷3)-sin2x的一个单调递增区间详解过程
人气:361 ℃ 时间:2020-03-23 04:05:44
解答
y=sin(2x-π/3)-sin2x
=2 cos[(2x-π/3+2x)/2] sin[(2x-π/3-2x)/2]
=2 cos(2x-π/6) sin(-π/6)
=- cos(2x-π/6)
=- cos(2x-π/6)
单调递增区间:
2kπ<=2x-π/6<=2kπ+3π/2
2kπ+π/6<=2x<=2kπ+3π/2+π/6
2kπ+π/6<=2x<=2kπ+5π/3
kπ+π/12<=x<=kπ+5π/6谢谢了单调递增区间:2kπ<=2x-π/6<=2kπ+π2kπ+π/6<=2x<=2kπ+π+π/62kπ+π/6<=2x<=2kπ+7π/6kπ+π/12<=x<=kπ+7π/12
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