y=sin(2x-π/3)的单调递增区间是?
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y=sin(2x-π/3)的单调递增区间是?
y=sin(2x-π/3)的单调递增区间是?
y=sin(2x-π/3)的单调递增区间是?
sin递增
所以2kπ-π/2
-π/2+2kπ<2x-π/3<π/2+2kπ
-π/6+2kπ<2x<5π/6+2kπ
-π/12+kπ
y=sin(2x-π/3)的单调递增区间是
2kPai-Pai/2<=2x-Pai/3<=2kPai+Pai/2
即有[kPai-Pai/12,kPai+5Pai/12]
由2kπ-π/2≤2X-π/3≤2kπ+π/2(k∈Z)得
kπ-π/12≤X≤kπ+5π/12(k∈Z),
Y=sin(2X-π/3)的递增区间是
[kπ-π/12,kπ+5π/12](k∈Z).
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令2kπ-π/2≤2x-π/3≤2kπ+π/2
解得 kπ-π/12≤x≤kπ+5π/12
即单调递增区间为[kπ-π/12,kπ+5π/12],k是整数
递增区间:2x-π/3,属于[2Kπ-π/2,2Kπ+π/2]
x在[Kπ-π/12,Kπ+5π/12]上单调递增
Y =罪(π/ 3-2X)=-SIN(2X-π/ 3)
即寻求Y =罪(2X-π/ 3)保存时间间隔
2kπ+π/ 2≤2倍-π/ 3≤2kπ+3π/ 2
2kπ+5π/ 6≤2×≤2kπ11π/ 6
Kπ5π/12≤X≤Kπ11π/12 BR />函数y = SIN(π/ 3-2X)的单调递增的间隔
Kπ+5π/12Kπ+11π/12],K∈Z
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