Question

求单调增区间

已知向量a=(2cosX.cos2X).b=(sinX.1).令f(x)=a*b.求f(x)的单调递增区间。

Answer 1

f(x)=2sinxcosx+cos2x=sin2x+cos2x=(√2)sin(2x+π/4)
∴f(π/2)=(√2)sin(π+π/4)=1
又∵x∈[-π/2，π/2]
∴x∈[-3π/8，π/8]，f(x)单调递增
∴x∈[kπ-3π/8，kπ+π/8]，k∈Z，f(x)单调递增

Answer 2

f(x)=2sinxcosx+cos2x=√2sin(2x+π/4)
2kπ-π/2≤2x+π/4≤2kπ+π/2
kπ-3π/8≤x≤kπ+π/8
所以单调增区间为[kπ-3π/8,kπ+π/8]