Question

y=sin（2x-π/3）的单调递增区间是？

Answer 1

y=3sin(π/6-3x)=-3sin(3x-π/6),x∈[-π/2,π/2]的单调递增区间即y=3sin(3x-π/6)x∈[-π/2,π/2]的单调递减区间,由2kπ
π/2<=3x-π/6<=2kπ
3π/2,2kπ/3
2π/9<=x<=2kπ/3
5π/9,
k=-1,0时-4π/9<=x<=-π/9，2π/9<=x<=5π/9，结合x∈[-π/2,π/2]得
y=3sin(π/6-3x)x∈[-π/2,π/2]的单调递增区间为：[-4π/9，-π/9]，[2π/9，π/2]

Answer 2

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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Answer 3

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

Answer 4

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

Answer 5

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]