Đáp án:
b. \(x \ne - \pi + \dfrac{{k\pi }}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:1 - \sin x > 0\\
\to 1 > \sin x\\
\Leftrightarrow \sin x \ne 1\\
\to x \ne \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)\\
b.DK:\left\{ \begin{array}{l}
\cos \left( {x + 2019\pi } \right) \ne 0\\
\sin \left( {x + 2019\pi } \right) \ne 0
\end{array} \right.\\
\to \sin 2\left( {x + 2019\pi } \right) \ne 0\\
\to 2\left( {x + 2019\pi } \right) \ne k\pi \\
\to x + 2019\pi \ne \dfrac{{k\pi }}{2}\\
\to x \ne - 2019\pi + \dfrac{{k\pi }}{2}\\
Hay:x \ne - \pi + \dfrac{{k\pi }}{2}\left( {k \in Z} \right)
\end{array}\)
