Question

设向量a=(sinx,cosx),b=(sinx,根号3sinx),x属于R,函数f(x)=a(a+2b).(1)求函数

Answer 1

f(x)=a(a+2b)=1+2(sin²x+√3sinxcosx)=1+(1-cos2x)+√3sin2x=2+√3sin2x-cos2x=2+2sin(2x+π/6)

∴由2kπ-π/2≤2x+π/6≤2kπ+π/2得,kπ-π/3≤x≤kπ+π/6

∴单调递增区间为[kπ-π/3,kπ+π/6]

2+2sin(2x+π/6)≥2

得,sin(2x+π/6)≥0

2kπ≤2x+π/6≤2kπ+π,解得kπ-π/12≤x≤kπ+5π/12

∴{x|kπ-π/12≤x≤kπ+5π/12}