Question

已知直线方程xsinθ-y+1=0,则直线的倾斜角的取值范围是?

Answer 1

已知xsinθ-y+1=0
∴k=-A/B=sinθ
∵k=tanα
∴tanα=sinθ
∵-1≤sinθ≤1
∴-1≤tanα≤1
∴kπ-π/4≤α≤kπ+π/4 (k∈Z)
又∵0≤α＜π
∴0≤α≤π/4
∴{α|0≤α≤π/4}