Đáp án:
$\begin{array}{l}
1)Dkxd:\dfrac{{1 - \cos x}}{{1 + \sin x}} \ge 0\\
\Leftrightarrow \dfrac{{\cos x - 1}}{{\sin x + 1}} \le 0\\
\Leftrightarrow do:\left\{ \begin{array}{l}
- 1 \le \sin x \le 1\\
- 1 \le \cos x \le 1
\end{array} \right.\\
\Leftrightarrow \sin x + 1\# 0\\
\Leftrightarrow \sin x\# - 1\\
\Leftrightarrow x\# - \dfrac{\pi }{2} + k2\pi \\
2)Dkxd:\left\{ \begin{array}{l}
\tan x\# 0\\
\cos x\# 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\sin x\# 0\\
\cos x\# 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x\# k\pi \\
x\# \dfrac{\pi }{2} + k\pi
\end{array} \right.\\
\Leftrightarrow x\# \dfrac{{k\pi }}{2}\left( {k \in Z} \right)\\
3)Dkxd:\sin \left( {{x^2} + 2x - 1} \right) \le 1\\
\Leftrightarrow {x^2} + 2x - 1 \le \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow {\left( {x + 1} \right)^2} \le \dfrac{{\pi + 4}}{2} + k2\pi \\
\Leftrightarrow - \sqrt {\dfrac{{\pi + 4}}{2} + k2\pi } - 1 \le x \le \sqrt {\dfrac{{\pi + 4}}{2} + k2\pi } - 1\\
4)Dkxd:\cos x\# 1\\
\Leftrightarrow x\# k2\pi \\
6)Dkxd:\left\{ \begin{array}{l}
\sin x\# 0\\
\cos 2x\# 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x\# k2\pi \\
2x\# \dfrac{\pi }{2} + k\pi
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x\# k2\pi \\
x\# \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}
\end{array} \right.\\
Vay\,x\# \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}
\end{array}$
