So kπ≦a/2≦kπ+π/4

Because of these two angles, the end edges will overlap only when their angles are different by an integer multiple of 2π.

Because k is an arbitrary real number, the value range of kπ is not clear, so we discuss it.

K=2n, that is, k is an even number, where 2nπ≦a/2≦2nπ+π/4 and a/2 are in the first quadrant.

K=2n+ 1, that is, k is an odd number. At this time, 2nπ+π≦a/2≦2nπ+5π/4, and a/2 is in the third quadrant.

So is the other one.

A/4 should be divided into k=4n, k = 4n+ 1, k = 4n+2 and k = 4n+3.

Teach you a simple method.

If you want to find the position of a/n, you can divide each quadrant n into equal parts and number it counterclockwise, 1234, 1234, ...

A is the quadrant angle, and a/n is in the range of logarithm.

If a is the first quadrant angle, then a/n is in the range corresponding to the number one.