Question

已知向量A=(2cos^2,根号3) 向量B=(1,sin2x),函数F(X)=向量a*b

求函数f(x)的单调递减区间

Answer 1

f(X)=向量a*b
=2cos²x+√3sin2x
=√3sin2x+cos2x+1
=2sin(2x+π/6)+1
由 2kπ+π/2≤2x+π/6≤2kπ+3π/2,k∈Z
得：kπ+π/6≤x≤kπ+2π/3,k∈Z
∴函数f(x)的单调递减区间：
[kπ+π/6,kπ+2π/3],k∈Z