1、最小正周期=2π/w=2π/2=π.
2、根据题意：
0<=x<=π/2
0<=2x<=π
π/6<=2x+π/6<=7π/6
所以：f(x)min=2sin(7π/6)=-1;
f(x)max=2sin(π/2)=2.
3.从-π到π，等同从0到2π。
根据题意:
2kπ+π/2<=2x+π/6<=2kπ+3π/2为递减区间
2kπ+π/3<=2x<=2kπ+4π/3
kπ+π/6<=x<=kπ+2π/3.
所以当k等于0时，有区间【π/6,2π/3];
当k等于1时有区间[7π/6,5π/3].
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