Question

已知函数y=sin^2x+sin2x+3cos^2x

已知函数y=sin^2x+sin2x+3cos^2x，求函数的最小值和单调减区间

Answer 1

y=(1-cos2x)/2+sin2x+3(1+cos2x)/2
=sin2x+cos2x+2
=√2sin(2x+π/4)+2
所以最小值=-√2+2

sin递减则2kπ+π/2<2x+π/4<2kπ+3π/2
2kπ+π/4<2x<2kπ+5π/4
kπ+π/8<x<kπ+5π/8
减区间(kπ+π/8，kπ+5π/8)