If β∈ [- 4 π, 0], and the terminal edge of β is the same as that of α in (1), then β is obtained

If β∈ [- 4 π, 0], and the terminal edge of β is the same as that of α in (1), then β is obtained

-1500°=﹣5×360º+300º
=﹣10π+5π／3
∵ β has the same terminal edge as α,
∴β=2kπ+5π／3,k∈Z,
∵β∈[-4π,0]
Ψ k = - 2 or - 1
Ψ β = - 7 π / 3 or - π / 3

Let 1480 ° be written in the form of 2K π + a (K ∈ Z, a ∈ [0,2 π]), if β ∈ [- 4 π, 0], and β is related to the terminal edge of α in the above question
Let 1480 ° be written in the form of 2K π + a (K ∈ Z, a ∈ [0,2 π]); if β ∈ [- 4 π, 0], and β is the same as the terminal edge of α in the above question, then find β

1480=8π+40
Because beta is negative, it should be - 320
-320/360=-16/9π
Minus 2 π for two answers
-16/9π ,-34/9π

The - 1485 degree is reduced to 2K π + α (0 ≤ α < 2 π, K ∈ z), and its quadrant is determined
-1485 degrees = - 33 / 4 π = - 10 π + 7 / 4 π
Why can't it be equal to - 9 π + 3 / 4

A period is 2 π, so it is even