求函数y=2sin(3x+π/4)的单调递增区间
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解:
y=2sin(3x+π/4)
依题意
-π/2+2kπ<=3x+π/4<=π/2+2kπ
-π/4+2kπ/3<=x<=π/12+2kπ/3
所以增区间为[-π/4+2kπ/3,π/12+2kπ/3]
y=2sin(3x+π/4)
依题意
-π/2+2kπ<=3x+π/4<=π/2+2kπ
-π/4+2kπ/3<=x<=π/12+2kπ/3
所以增区间为[-π/4+2kπ/3,π/12+2kπ/3]
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