Đáp án:
$\begin{array}{l}
a)\sqrt 2 .\sin \left( {x + \dfrac{\pi }{3}} \right) + 1 = 0\\
\Leftrightarrow \sin \left( {x + \dfrac{\pi }{3}} \right) = - \dfrac{1}{{\sqrt 2 }}\\
\Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{3} = - \dfrac{\pi }{4} + k2\pi \\
x + \dfrac{\pi }{3} = \pi + \dfrac{\pi }{4} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{{7\pi }}{{12}} + k2\pi \\
x = \dfrac{{11\pi }}{{12}} + k2\pi
\end{array} \right.\\
Vậy\,\left[ \begin{array}{l}
x = - \dfrac{{7\pi }}{{12}} + k2\pi \\
x = \dfrac{{11\pi }}{{12}} + k2\pi
\end{array} \right.\\
c)3\cot \left( {2x - \dfrac{\pi }{4}} \right) + \sqrt 3 = 0\\
\Leftrightarrow \cot \left( {2x - \dfrac{\pi }{4}} \right) = - \dfrac{{\sqrt 3 }}{3}\\
\Leftrightarrow 2x - \dfrac{\pi }{4} = \dfrac{\pi }{3} + k\pi \\
\Leftrightarrow 2x = \dfrac{{7\pi }}{{12}} + k\pi \\
\Leftrightarrow x = \dfrac{{7\pi }}{{24}} + \dfrac{{k\pi }}{2}\\
Vậy\,x = \dfrac{{7\pi }}{{24}} + \dfrac{{k\pi }}{2}
\end{array}$
