Task: Solve the inequality $$-2\cdot\sin \frac{x}{3}>1$$ for $x\in\mathbb{R}$. Marek solved the task using the following steps:
(1) He applied equivalent transformations to convert the inequality to the form: $$\sin\frac{x}{3}<-\frac12$$ (2) By substituting $\frac x3=a$, he obtained the inequality: $$\sin a<-\frac12$$ (3) He solved the equation $\sin a=-\frac12$∶ $$a_1=\frac{7\pi}{6}+k\cdot2\pi\ \mbox{ and } a_2=\frac{11\pi}{6}+k\cdot2\pi,\ \mbox{ for }k\in\mathbb{Z}$$ (4) With the aid of the unit circle, he solved the inequality $\sin a<-\frac12$:
$$a\in\bigcup_{k\in\mathbb{Z}}\left(\frac{7\pi}{6}+k\cdot2\pi;\frac{11\pi}{6}+k\cdot2\pi\right),\ \mbox{ for }k\in\mathbb{Z}$$
(5) He substituted the obtained values $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$ back, expressed the unknown $x$, and added the period:
$$K=\bigcup_{k\in\mathbb{Z}}\left(\frac{7\pi}{2}+k\cdot2\pi;\frac{11\pi}{2}+k\cdot2\pi\right),\ \mbox{ for }k\in\mathbb{Z}$$
The result is not correct. In which step did Marek make an error?
The error is in step (1). Marek incorrectly changed the inequality sign to the opposite.
The error is in step (2). The substitution $3x=a$ should have been used.
The error is in step (3). Marek incorrectly solved the equation $\sin a=-\frac12$.
The error is in step (4). Marek incorrectly solved the inequality $\sin a<-\frac12$. He should have chosen the complementary part of the unit circle, i.e., $$a\in\bigcup_{k\in\mathbb{Z}}\left(-\frac{\pi}{6}+k\cdot2\pi;\frac{7\pi}{6}+k\cdot2\pi\right)$$
The error is in step (5). Marek expressed the solution $K$ incorrectly.
Let’s show the correct procedure. By solving the inequality $\sin a<-\frac12$ in step (4), we obtain: $$a\in\bigcup_{k\in\mathbb{Z}}\left(\frac{7\pi}{6}+k\cdot2\pi;\frac{11\pi}{6}+k\cdot2\pi\right) ,\ \mbox{ for }k\in\mathbb{Z}$$ Then, we substitute the values $a_1=\frac{7\pi}{6}+k\cdot2\pi$ and $a_2=\frac{11\pi}{6}+k\cdot2\pi$ back into the substitution $\frac{x}{3}=a$ and obtain two equations. From these, we determine the endpoints of the intervals that are solutions to the inequality $\sin\frac{x}{3}<-\frac12$: \begin{aligned} &\frac{x_1}{3}=\frac{7\pi}{6}+k\cdot2\pi\Rightarrow x_1=\frac{7\pi}{2}+k\cdot6\pi,\ \mbox{ for }k\in\mathbb{Z}\cr &\frac{x_2}{3}=\frac{11\pi}{6}+k\cdot2\pi\Rightarrow x_2=\frac{11\pi}{2}+k\cdot6\pi,\ \mbox{ for } k\in\mathbb{Z} \end{aligned} Therefore, the solution to the given inequality is: $$x\in\bigcup_{k\in\mathbb{Z}}\left(\frac{7\pi}{2}+k\cdot6\pi;\frac{11\pi}{2}+k\cdot6\pi\right),\ \mbox{ for }k\in\mathbb{Z}$$
