Solve the inequality: $$\tan\left(x-\frac{3\pi}{8}\right)>1\ \mbox{ for }x\in\mathbb{R}$$ Michael solved the task in the following steps:
(1) Using the substitution $m=x-\frac{3\pi}{8}$, he rewrote the inequality in the form: $$\tan m>1$$ (2) He then solved the equation $\tan m=1$: $$m=\frac{\pi}{4}+k\cdot\pi,\ \mbox{ for }k\in\mathbb{Z}$$ (3) He expressed the result in terms of the unknown $x$, obtaining the solution of the equation $\tan\left(x-\frac{3\pi}{8}\right)=1$: $$x-\frac{3\pi}{8}=\frac{\pi}{4}+k\cdot\pi\Rightarrow x=\frac{5\pi}{8}+k\cdot\pi,\ \mbox{ for }k\in\mathbb{Z}$$ (4) Next, he determined the points where the function $\tan\left(x-\frac{3\pi}{8}\right)$ is not defined: $$\bigcup_{k\in\mathbb{Z}}\left\{\frac{7\pi}{8}+k\cdot\pi\right\}$$ (5) He stated that the tangent function is increasing, so the value of the function $\tan\left(x-\frac{3\pi}{8}\right)$ will be greater than $1$ when $x$ is greater than $\frac{5\pi}{8}$. By combining this result with the result obtained in step (4), he concluded: $$\tan\left(x-\frac{3\pi}{8}\right)>1\Leftrightarrow x >\frac{5\pi}{8}\ \mbox{ and } x\neq\frac{7\pi}{8}+k\cdot\pi,\ \mbox{ for }k\in\mathbb{Z}$$
Thus, he stated that the solution to the given inequality is: $$K=\left(\frac{5\pi}{8};+\infty\right)\backslash\bigcup_{k\in\mathbb{Z}}\left\{\frac{7\pi}{8}+k\cdot \pi\right\}$$ The solution is not correct. In which step did Michal make an error?
The error is in step (1). In this case, a substitution cannot be used.
The error is in step (2). One solution of the equation $\tan m=1$ is $m=\frac{\pi}{4}$ and the basic period of the function $\tan(m)$ is $2\pi$. Therefore, all solutions are: $$m=\frac{\pi}{4}+2\cdot k\cdot\pi,\ \mbox{ for }k\in\mathbb{Z}$$
The error is in step (3). The unknown $x$ is not expressed correctly. It should be: $$x=-\frac{\pi}{8}+k\cdot\pi,\ \mbox{ for }k\in\mathbb{Z}$$
The error is in step (4). The function is not defined for: $$\bigcup_{k\in\mathbb{Z}}\left\{\frac{3\pi}{8}+k\cdot\pi\right\}$$
The error is in step (5). The function is not increasing over the entire domain. The tangent function is increasing only between every two adjacent points where it is not defined.
Let’s present the correct solution. The function is not defined at the points:
$$\bigcup_{k\in\mathbb{Z}}\left\{\frac{7\pi}{8}+k\cdot\pi\right\}$$
The tangent function is periodic and increasing between every two adjacent points where it is not defined. Therefore:
$$\tan\left(x-\frac{3\pi}{8}\right)>1\Leftrightarrow x \in\left(\frac{5\pi}{8}+k\cdot\pi; \frac{7\pi}{8}+k\cdot\pi\right),\ \mbox{ for } k\in\mathbb{Z}$$
The solution of the inequality $\tan\left(x-\frac{3\pi}{8}\right)>1$ can finally by written as:
$$K=\bigcup_{k\in\mathbb{Z}}\left(\frac{5\pi}{8}+k\cdot\pi; \frac{7\pi}{8}+k\cdot\pi\right)$$