$\begin{array}{l} A = \sin 2x + \cos 2x + 3\\ A = \sqrt 2 .\left( {\dfrac{1}{{\sqrt 2 }}\sin 2x + \dfrac{1}{{\sqrt 2 }}\cos 2x} \right) + 3\\ A = \sqrt 2 \sin \left( {2x + \dfrac{\pi }{4}} \right) + 3\\ Do\, - 1 \le \sin \left( {2x + \dfrac{\pi }{4}} \right) \le 1\\ \to 3 - \sqrt 2 \le A \le 3 + \sqrt 2 \\ \to \left\{ \begin{array}{l} a = 3\\ b = 1 \end{array} \right. \Rightarrow a + b = 4 \to A \end{array}$
Dấu bằng xảy ra khi và chỉ khi: $\sin (2x+\dfrac{\pi}{4})=1\Leftrightarrow 2x+\dfrac{\pi}4=\dfrac{\pi}2+k2\pi$
$\Leftrightarrow x=\dfrac{\pi}8+k\pi$
