Đáp án:
$\begin{array}{l}
\cos \dfrac{1}{x} < 0\\
\Leftrightarrow \cos \dfrac{1}{x} < \cos \left( {\dfrac{\pi }{2} + k\pi } \right)\\
\Leftrightarrow \dfrac{\pi }{2} + k2\pi < \dfrac{1}{x} < \dfrac{{3\pi }}{2} + k2\pi \\
\Leftrightarrow \dfrac{{\pi + k4\pi }}{2} < \dfrac{1}{x} < \dfrac{{3\pi + k4\pi }}{2}\\
\Leftrightarrow \dfrac{2}{{3\pi + k4\pi }} < x < \dfrac{2}{{\pi + k4\pi }}\\
Vậy\,\dfrac{2}{{3\pi + k4\pi }} < x < \dfrac{2}{{\pi + k4\pi }}
\end{array}$
