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}