已知函数f(x)=-2sin(2x+φ)(|φ|

1个回答

  • 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|