| ||
2 |
1 |
2 |
| ||
2 |
1 |
2 |
=2(
| ||
2 |
1 |
2 |
=2sin(ωx−
π |
6 |
由−1≤sin(ωx−
π |
6 |
π |
6 |
可知函数f(x)的值域为[-3,1].
(II)由题设条件及三角函数图象和性质可知,y=f(x)的周期为π,
又由ω>0,得
2π |
ω |
于是有f(x)=2sin(2x−
π |
6 |
再由2kπ−
π |
2 |
π |
6 |
π |
2 |
解得kπ−
π |
6 |
π |
3 |
所以y=f(x)的单调增区间为[kπ−
π |
6 |
π |
3 |
π |
6 |
π |
6 |
ωx |
2 |
π |
2 |
| ||
2 |
1 |
2 |
| ||
2 |
1 |
2 |
| ||
2 |
1 |
2 |
π |
6 |
π |
6 |
π |
6 |
2π |
ω |
π |
6 |
π |
2 |
π |
6 |
π |
2 |
π |
6 |
π |
3 |
π |
6 |
π |
3 |