Answer:
[tex]\cot(x) = -1[/tex] whenever [tex]\displaystyle x = k\, \pi-\frac{\pi}{4}[/tex] radians, where [tex]k[/tex] could be any integer ([tex]k \in \mathbb{Z}[/tex], which includes positive whole numbers, negative whole numbers, and zero.)
Step-by-step explanation:
[tex]x = 45^\circ[/tex] (as in isoscele right triangles) would ensure that [tex]\displaystyle \cot(x) = 1[/tex]. Since cotangent is an odd function, [tex]\cot(-45^\circ) = -1[/tex].
Equivalently, when the angles are expressed in radians, [tex]\cot(-\pi / 4) = -1[/tex].
The cycle of cotangent is [tex]\pi[/tex] (or equivalently, [tex]180^\circ[/tex].) Therefore, if [tex]k[/tex] represents an integer, adding [tex]k\, \pi[/tex] to the input to cotangent would not change the output. In other words:
[tex]\displaystyle \cot\left(k\, \pi - \frac{\pi}{4}\right) = \cot(-\pi / 4) = -1[/tex].
Hence, [tex]\displaystyle x = k\, \pi-\frac{\pi}{4}[/tex] would be a solution to [tex]\cot(x) = -1[/tex] whenever [tex]k[/tex] is an integer.
Since [tex](-\pi / 4)[/tex] is the only solution to this equation in the period [tex](0,\, \pi)[/tex], all real solutions to this equation would be in the form [tex]\displaystyle x = k\, \pi-\frac{\pi}{4}[/tex] (where [tex]k[/tex] is an integer.)
