$\begin{array}{l} a)\sin x + \cos x = \sqrt 2 \\ \Leftrightarrow \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = \sqrt 2 \\ \Leftrightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = 1\\ \Leftrightarrow x + \dfrac{\pi }{4} = \dfrac{\pi }{2} + k2\pi \\ \Leftrightarrow x = \dfrac{\pi }{4} + k2\pi (k\in \mathbb Z)\\ b)\cos x - \sqrt 3 \sin x = 1\\ \Leftrightarrow 2\left( {\dfrac{1}{2}\cos x - \dfrac{{\sqrt 3 }}{2}\sin x} \right) = 1\\ \Leftrightarrow 2\cos \left( {x + \dfrac{\pi }{3}} \right) = 1\\ \Leftrightarrow \cos \left( {x + \dfrac{\pi }{3}} \right) = \dfrac{1}{2}\\ \Leftrightarrow \left[ \begin{array}{l} x + \dfrac{\pi }{3} = \dfrac{\pi }{3} + k2\pi \\ x + \dfrac{\pi }{3} = - \dfrac{\pi }{3} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = k2\pi \\ x = - \dfrac{{2\pi }}{3} + k2\pi \end{array} \right.\left( {k \in \mathbb Z} \right)\\ c)3\sin 2x + 4\cos 2x = 5\\ \Leftrightarrow 5\left( {\dfrac{3}{5}\sin 2x + \dfrac{4}{5}\cos 2x} \right) = 5\\ \Leftrightarrow 5\sin \left( {2x + \alpha } \right) = 5\left( {\alpha = \arccos \dfrac{3}{5}} \right)\\ \Leftrightarrow \sin \left( {2x + \alpha } \right) = 1\\ \Leftrightarrow 2x + \alpha = \dfrac{\pi }{2} + k2\pi \\ \Leftrightarrow x = \dfrac{\pi }{4} - \dfrac{\alpha }{2} + k\pi \left( {k \in \mathbb Z} \right) \end{array}$
