Question

已知函数f(x)=sin2x＋√3cos2x,x∈R (1)求f(x)的最小周期数 (2

求f(x)的单调递增区间

Answer 1

f(x)=sin2x＋√3cos2x
=2（1/2sin2x+ √3/2cos2x)
=2sin(2x+π/3)
T=2π/2=π
当-π/2+2kπ≤2x+π/3≤π/2+2kπ即
-5π/12+kπ≤x≤π/12+kπ
则f(x)单调递增区间为[-5π/12+kπ，π/12+kπ](
k∈z)

Answer 2

解：（1）f(x)=sin2x+√3cos2x=2sin(2x+π/3)
故T=2π/2=π
（2）由-π/2+2kπ<=2x+π/3<=π/2+2kπ
得，-5π/12+kπ<=x<=π/12+kπ
故递增区间是[-5π/12+kπ,π/12+kπ] k为整数

Answer 3

Answer 4

f(x)=2sin(2x+л/3)
所以
（1）T=2л /w=2л /2=л
（2）只需要-л/2+2kл<=2x+л/3 ≤л/2+2kл
所以解出-5/12л+kл<=x ≤л/12+kл