Question

设函数f(x)=sin(ωx+φ).(ω＞0，-π/2＜π＜π/2)的图象关于直线x=2π/3对称

设函数f(x)=sin(ωx+φ).(ω＞0，-π/2＜π＜π/2)的图象关于直线x=2π/3对称.它的周期是π.则y=f(x)的单调递增区间是
这道题很简单的，不要想复杂了。第二个条件是(ω＞0，-π/2＜φ＜π/2)

Answer 1

周期是π, ω = 2
x = 2π/3, 2x+φ = π/2为对称轴，φ = π/2 - 2*2π/3 = -5π/6
f(x) = sin(2x - 5π/6)
2x - 5π/6 = -π/2, x = π/6
2x - 5π/6 = π/2, x = 2π/3
单递增区间: (π/6 + kπ, 2π/3 + kπ)