tan(x/2) = sin(x/2)
sin(x/2) / cos(x/2) = sin(x/2)
sin(x/2) / cos(x/2) - sin(x/2) = 0
sin(x/2) (1/cos(x/2) - 1) = 0
Then either
sin(x/2) = 0 or 1/cos(x/2) - 1 = 0
• If sin(x/2) = 0, we have
x/2 = arcsin(0) + 2nπ or x/2 = π - arcsin(0) + 2nπ
(where n is any integer)
x/2 = 2nπ or x/2 = π + 2nπ
x = 4nπ or x = 2π + 4nπ
• If 1/cos(x/2) - 1 = 0, we have
1/cos(x/2) = 1
cos(x/2) = 1
x/2 = arccos(1) + 2nπ
x/2 = 2nπ
x = 4nπ
but we already have this family of solutions accounted for.
