$\mathrm{cot}\,x < -1$

Project ID: 
3000000086
Question: 

Solve the inequality: $$\cot ⁡x<-1\quad\mbox{for }x\in\mathbb{R}$$

Robert solved the task in the following steps:

(1) He determined the points at which the function $y = \cot x$ is not defined and, thus, determined its domain: $$D=\mathbb{R}\backslash\left\{k\cdot\pi; k\in\mathbb{Z}\right\}$$

(2) Robert found the solution of the equation $\cot ⁡x=-1$, which is the set: $$K_1=\left\{\frac{3\pi}{4}+k\cdot\pi;k\in\mathbb{Z}\right\}$$

(3) He stated that the cotangent function is decreasing over its domain, so the value of $\cot x$ will be less than $-1$ when $x$ is greater than $\frac{3\pi}{4}$. Thus, he wrote: $$\cot ⁡x<-1\Leftrightarrow x > \frac{3\pi}{4}$$

(4) Finally, he excluded points that are not in the domain of $\cot x$ from the result obtained in step (3) and wrote down the solution: $$K=\left(\frac{3\pi}{4};+\infty\right)\backslash \bigcup_{k\in\mathbb{Z}}\left\{k\cdot\pi\right\}$$ The solution is not correct. In which step did Robert make a mistake?

Answer 1: 

The mistake is in step (1). The domain of the cotangent function is not specified correctly.

Answer 2: 

The mistake is in step (2). The period in the solution of the equation $\cot ⁡x=-1$ is determined incorrectly.

Answer 3: 

The mistake is in step (3). The function $\cot x$ is not decreasing over the entire domain.

Answer 4: 

The mistake is in step (4). Points in which the function is not defined do not need to be excluded from the interval obtained in step (3).

Fixed Answer: 
All Fixed
Correct Answer: 
Answer 3
Hint: 

Let’s present the correct solution. The function $\cot x$ is decreasing only on open intervals bounded by two adjacent points where the cotangent is not defined, i.e., on the intervals: $$\ldots,(-\pi;0),\ (0;\pi),\ (\pi;2\pi),\ (2\pi;3\pi),\ (3\pi;4\pi),\ldots$$ Therefore, we can find the solution to the inequality $\cot ⁡x<-1$ on each interval $(0+k\cdot\pi;\pi+k\cdot\pi)$ as: $$\frac{3\pi}{4}+k\cdot\pi<x<\pi+k\cdot\pi,\ \mbox{ where } k\in\mathbb{Z},$$ The solution can be written as the union of all the corresponding subintervals: $$K=\bigcup_{k\in\mathbb{Z}}\left(\frac{3\pi}{4}+k\cdot\pi;\pi+k\cdot\pi\right)$$