f(x)=-2sin(2x+φ)的单调递增区间为[2kπ+π/2,2kπ+3π/2]
即2kπ+π/2≤2x+φ≤2kπ+3π/2
解得kπ+π/4-φ/2≤x≤kπ+3π/4-φ/2
因为(π/5,5π/8)是f(x)的一个单调递增区间
所以(π/5,5π/8)包含于[kπ+π/4-φ/2,kπ+3π/4-φ/2]
即kπ+π/4-φ/2≤π/5,5π/8≤kπ+3π/4-φ/2
解得2kπ+π/10≤φ≤2kπ+π/4
又因为|φ|
f(x)=-2sin(2x+φ)的单调递增区间为[2kπ+π/2,2kπ+3π/2]
即2kπ+π/2≤2x+φ≤2kπ+3π/2
解得kπ+π/4-φ/2≤x≤kπ+3π/4-φ/2
因为(π/5,5π/8)是f(x)的一个单调递增区间
所以(π/5,5π/8)包含于[kπ+π/4-φ/2,kπ+3π/4-φ/2]
即kπ+π/4-φ/2≤π/5,5π/8≤kπ+3π/4-φ/2
解得2kπ+π/10≤φ≤2kπ+π/4
又因为|φ|