Question

Find all solutions
for cotx=-1​

Answer 1

`\cot(x) = -1` whenever
`\displaystyle x = k\, \pi-(\pi)/(4)` radians, where
`k` could be any integer (
`k \in \mathbb{Z}`, which includes positive whole numbers, negative whole numbers, and zero.)

Explanation:

`x = 45^\circ` (as in isoscele right triangles) would ensure that
`\displaystyle \cot(x) = 1`. Since cotangent is an odd function,
`\cot(-45^\circ) = -1`.

Equivalently, when the angles are expressed in radians,
`\cot(-\pi / 4) = -1`.

The cycle of cotangent is
`\pi` (or equivalently,
`180^\circ`.) Therefore, if
`k` represents an integer, adding
`k\, \pi` to the input to cotangent would not change the output. In other words:

`\displaystyle \cot\left(k\, \pi - (\pi)/(4)\right) = \cot(-\pi / 4) = -1`.

Hence,
`\displaystyle x = k\, \pi-(\pi)/(4)` would be a solution to
`\cot(x) = -1` whenever
`k` is an integer.

Since
`(-\pi / 4)` is the only solution to this equation in the period
`(0,\, \pi)`, all real solutions to this equation would be in the form
`\displaystyle x = k\, \pi-(\pi)/(4)` (where
`k` is an integer.)