Question

高一数学！

Answer 1

答：
f(x)=2sin²x+2√3sinxcosx-1
=1-cos2x+√3sin2x-1
=2*[(√3/2)sin2x-(1/2)cos2x]
=2sin(2x-π/6)

（1）f(x)=2sin(2x-π/6)的单调增区间满足：
2kπ-π/2<=2x-π/6<=2kπ+π/2
所以：单调增区间为[kπ-π/6，kπ+π/3]，k∈Z

（2）f(x)=2sin(2x-π/6)=2sin[2(x-π/12)]可以由y=sinx向右平移π/12个单位后横坐标缩短1/2，然后纵坐标伸长2倍得到。