已知函数f(x)=sin(2x+7π/4)+cos(2x-3π/4),x属于R.1,求f(x)的单调递增区间和对称轴方程,2,求f(x)>0使得自变量x的集合,最I好把过程写一遍.

Question

已知函数f(x)=sin(2x+7π/4)+cos(2x-3π/4),x属于R.
1,求f(x)的单调递增区间和对称轴方程,2,求f(x)>0使得自变量x的集合,最I好把过程写一遍.

Answer 1

f(x)=sin(2x+7π/4)+cos(2x-3π/4)=2sin(2x-π/4)
由-π/2+2kπ≤2x-π/4≤π/2+2kπ,k∈Z,
得-π/8+kπ≤x≤3π/8+kπ,k∈Z,
即f(x)的单增区间是[-π/8+kπ,3π/8+kπ],k∈Z.
对称轴方程是x=3π/8+kπ,k∈Z.
∵f(x)=2sin(2x-π/4)>0
       ∴2kπ<2x-π/4<π+2kπ,k∈Z,
       故x的集合是{x|π/8+kπ