设函数F(X)=SIN(2X+Φ) (-π
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设函数F(X)=SIN(2X+Φ) (-π
设函数F(X)=SIN(2X+Φ) (-π
设函数F(X)=SIN(2X+Φ) (-π
由于函数关于这条直线对称,
根据此函数特点可以确定函数在x=π/8这一点f(x)=1或者-1,
所以2x+φ=kπ+π/2(k为实整数),
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.
原方程为f(x)=sin(2x-3π/4),sinx的单调递增区间是[2kπ-π/2,2kπ+π/2],
所以2kπ-π/2<2x-3π/4<2kπ+π/2,
由此解得kπ-π/8
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