$\begin{array}{l}
t = \sin x + \cos x \Rightarrow {t^2} = {\left( {\sin x + \cos x} \right)^2} = 1 + \sin 2x\left( {t \in \left[ { - \sqrt 2 ;\sqrt 2 } \right]} \right)\\
\Leftrightarrow \sin 2x = {t^2} - 1\\
2\left( {\sin x + \cos x} \right) + 3\sin 2x = 2\\
\Leftrightarrow 2t + 3\left( {{t^2} - 1} \right) = 2\\
\Leftrightarrow 3{t^2} + 2t - 5 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
t = 1\\
t = - \dfrac{5}{3}(L)
\end{array} \right.\\
\Rightarrow \sin x + \cos x = 1 \Rightarrow \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = 1\\
\Leftrightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{4} = \dfrac{\pi }{4} + k2\pi \\
x + \dfrac{\pi }{4} = \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = k2\pi \\
x = \dfrac{\pi }{2} + k2\pi
\end{array} \right.\left( {k \in \mathbb{Z}} \right)\\
b)\left( {1 - \sqrt 2 } \right)\left( {1 + \sin x + \cos x} \right) = \sin 2x\\
\Leftrightarrow \left( {1 - \sqrt 2 } \right)\left( {1 + t} \right) = {t^2} - 1\\
\Leftrightarrow \left( {1 - \sqrt 2 } \right)t + \left( {1 - \sqrt 2 } \right) = {t^2} - 1\\
\Leftrightarrow {t^2} + \left( {\sqrt 2 - 1} \right)t + \sqrt 2 - 2 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
t = - 1\\
t = 2 - \sqrt 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = - 1\\
\sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = 2 - \sqrt 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin \left( {x + \dfrac{\pi }{4}} \right) = - \dfrac{{\sqrt 2 }}{2}\\
\sin \left( {x + \dfrac{\pi }{4}} \right) = \sqrt 2 - 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{4} = - \dfrac{\pi }{4} + k2\pi \\
x + \dfrac{\pi }{4} = \dfrac{{5\pi }}{4} + k2\pi \\
x + \dfrac{\pi }{4} = \arcsin \left( {\sqrt 2 - 1} \right) + k2\pi \\
x - \dfrac{\pi }{4} = \pi - \arcsin \left( {\sqrt 2 - 1} \right) + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{ - \pi }}{2} + k2\pi \\
x = \pi + k2\pi \\
x = - \dfrac{\pi }{4} + \arcsin \left( {\sqrt 2 - 1} \right) + k2\pi \\
x = \dfrac{{5\pi }}{4} - - \arcsin \left( {\sqrt 2 - 1} \right) + k2\pi
\end{array} \right.\left( {k \in \mathbb{Z}} \right)\\
c)\left( {1 + \sqrt 2 } \right)\left( {\sin x + \cos x} \right) - \sin 2x = 1 + \sqrt 2 \\
\Leftrightarrow \left( {1 + \sqrt 2 } \right)t - \left( {{t^2} - 1} \right) = 1 + \sqrt 2 \\
\Leftrightarrow {t^2} - \left( {1 + \sqrt 2 } \right)t + \sqrt 2 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
t = 1\\
t = \sqrt 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = 1\\
\sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = \sqrt 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin \left( {x + \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2}\\
\sin \left( {x + \dfrac{\pi }{4}} \right) = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k2\pi \\
x = k2\pi \\
x = \dfrac{\pi }{2} + k2\pi
\end{array} \right.\left( {k \in \mathbb{Z}} \right)
\end{array}$
