Đáp án:
$\begin{array}{l}
a)\sin x + \cos x = - 1\\
\Leftrightarrow \sqrt 2 .\sin \left( {x + \dfrac{\pi }{4}} \right) = - 1\\
\Leftrightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = - \dfrac{1}{{\sqrt 2 }}\\
\Leftrightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = \sin \left( { - \dfrac{\pi }{4}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{4} = - \dfrac{\pi }{4} + k2\pi \\
x + \dfrac{\pi }{4} = \pi + \dfrac{\pi }{4} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{\pi }{2} + k2\pi \\
x = \pi + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
Vậy\,x = - \dfrac{\pi }{2} + k2\pi ;x = \pi + k2\pi \left( {k \in Z} \right)\\
b){\sin ^2}x - 4\sin x + 3 = 0\\
\Leftrightarrow \left( {\sin x - 3} \right)\left( {\sin x - 1} \right) = 0\\
\Leftrightarrow \sin x = 1\left( {do: - 1 \le \sin x \le 1} \right)\\
\Leftrightarrow x = \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)\\
Vậy\,x = \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)
\end{array}$