由于函數(shù)關(guān)于這條直線對(duì)稱,
根據(jù)此函數(shù)特點(diǎn)可以確定函數(shù)在x=π/8這一點(diǎn)f(x)=1或者-1,
所以2x+φ=kπ+π/2(k為實(shí)整數(shù)）,
即π/4+φ=kπ+π/2,
所以φ=kπ+π/4,
又-π<φ<0,所以k只能取-1,
所以φ=-3π/4.
原方程為f(x)=sin(2x-3π/4),sinx的單調(diào)遞增區(qū)間是[2kπ-π/2,2kπ+π/2],
所以2kπ-π/2<2x-3π/4<2kπ+π/2,
由此解得kπ-π/8即f(x)的單調(diào)區(qū)間為[kπ-π/8,kπ+5π/8],k為實(shí)整數(shù).