Question

已知函数f(x)=sin(2x-π/6),x∈R

（1）求f（x）的最小正周期T
（2）求f（0）的值
（3）设α是第一象限角，且f（α+π/3）=3/5，求sinα的值

Answer 1

解：
f(x)=sin2xcosπ/6+cos2xsinπ/6+sin2xcosπ/6-cos2xsinπ/6+1+cos2x
=2sin2xcosπ/6+cos2x+1
=√3sin2x+cos2x+1
=2sin(2x+π/6)+1
⑴f(x)取得最大值3，此时2x+π/6=π/2+2kπ，即x=π/6+kπ，k∈Z
故x的取值集合为{x|x=π/6+kπ，k∈Z}
⑵由2x+π/6∈[-π/2+2kπ,π/2+2kπ](k∈Z)得，
x∈[-π/3+kπ,π/6+kπ](k∈Z)
故函数f(x)的单调递增区间为[-π/3+kπ,π/6+kπ](k∈Z)
⑶f(x)≥2⇔2sin(2x+π/6)+1≥2⇔sin(2x+π/6)≥1/2
⇔π/6+2kπ≤2x+π/6≤5π/6+2kπ
⇔kπ≤x≤π/3+kπ(k∈Z)
故f(x)≥2的x的取值范围是[kπ,π/3+kπ](k∈Z)