Lời giải

\[1 + \sin x + \cos x = 2\cos \left( {\frac{x}{2} - \frac{\pi }{4}} \right)\]

\( \Leftrightarrow 1 + \sin x + \cos x = 2\left( {\cos \frac{x}{2}.\cos \frac{\pi }{4} + \sin \frac{x}{2}.\sin \frac{\pi }{4}} \right)\)

\( \Leftrightarrow {\left( {\cos \frac{x}{2} + \sin \frac{x}{2}} \right)^2} + \left( {{{\cos }^2}\frac{x}{2} - {{\sin }^2}\frac{x}{2}} \right) = \sqrt 2 \left( {\cos \frac{x}{2} + \sin \frac{x}{2}} \right)\)

\( \Leftrightarrow {\left( {\cos \frac{x}{2} + \sin \frac{x}{2}} \right)^2} + \left( {\cos \frac{x}{2} + \sin \frac{x}{2}} \right)\left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right) - \sqrt 2 \left( {\cos \frac{x}{2} + \sin \frac{x}{2}} \right) = 0\)

\[ \Leftrightarrow \left( {\cos \frac{x}{2} + \sin \frac{x}{2}} \right)\left( {\cos \frac{x}{2} + \sin \frac{x}{2} + \cos \frac{x}{2} - \sin \frac{x}{2} - \sqrt 2 } \right) = 0\]

\[ \Leftrightarrow \sqrt 2 \sin \left( {\frac{x}{2} + \frac{\pi }{4}} \right)\left( {2\cos \frac{x}{2} - \sqrt 2 } \right) = 0\]

\[ \Leftrightarrow \left[ \begin{array}{l}\sin \left( {\frac{x}{2} + \frac{\pi }{4}} \right) = 0\\\cos \frac{x}{2} = \frac{{\sqrt 2 }}{2}\end{array} \right.\]

\[ \Leftrightarrow \left[ \begin{array}{l}\frac{x}{2} + \frac{\pi }{4} = k\pi \\\frac{x}{2} = \pm \frac{\pi }{4} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\]

\[ \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{2} + k2\pi \\x = \pm \frac{\pi }{2} + k4\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\]

\[ \Leftrightarrow x = \frac{\pi }{2} + k\pi \,\,\,\,\left( {k \in \mathbb{Z}} \right)\]

Vậy phương trình đã cho có nghiệm là \[x = \frac{\pi }{2} + k\pi \,\,\,\,\left( {k \in \mathbb{Z}} \right)\].