Question

函数y=1＋4cos²x的单调递增区间是

Answer 1

y=cos(π/4-2x)=cos(2x-π/4)
2kπ-π<=2x-π/4<=2kπ，则kπ-3π/8<=x<=kπ+π/8
所以，函数的单调递增区间是[kπ-3π/8,kπ+π/8]

Answer 2

y=1＋4cos²x=1+2(1+cos2x)=3+2cos2x
cos2x在[2kπ-π,2kπ]上单调递增区
cosx在[kπ-π/2,kπ]上单调递增区