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$2\sqrt 2 \left( {\sin x + \cos x} \right)\cos x = 3 + \cos 2x$$\Leftrightarrow \sqrt 2 \sin x + \left( {\sqrt 2 - 1} \right)\cos 2x = 3 - \sqrt 2$ Có: $\left\{ \begin{array} {a^2} + {b^2} = {\left( {\sqrt 2 } \right)^2} + {\left( {\sqrt 2 - 1} \right)^2} = 5 - 2\sqrt 2 \\ {c^2} = {\left( {3 - \sqrt 2 } \right)^2} = 11 - 6\sqrt 2 \\ \end{array} \right.$ Ta sẽ chứng minh: ${a^2} + {b^2}{\text{ < }}{{\text{c}}^{\text{2}}}$$\Leftrightarrow 5 - 2\sqrt 2 {\text{ < 11 - 6}}\sqrt {\text{2}}$$\Leftrightarrow 4\sqrt 2 {\text{ < 6}} \Leftrightarrow {\left( {{\text{4}}\sqrt {\text{2}} } \right)^2}{\text{ < }}{{\text{6}}^{\text{2}}}$$\Leftrightarrow 32{\text{ < 36}}$(đúng) Vậy phương trình vô nghiệm.

$2\sqrt 2 \left( {\sin x + \cos x} \right)\cos x = 3 + \cos 2x$$ \Leftrightarrow \sqrt 2 \sin 2x + \left( {\sqrt 2 - 1} \right)\cos 2x = 3 - \sqrt 2 $Có:$\left\{ $$\begin{array} {a^2} + {b^2} = {\left( {\sqrt 2 } \right)^2} + {\left( {\sqrt 2 - 1} \right)^2} = 5 - 2\sqrt 2 \\ {c^2} = {\left( {3 - \sqrt 2 } \right)^2} = 11 - 6\sqrt 2 \\ \end{array}$$ \right.$ Ta sẽ chứng minh: ${a^2} + {b^2}{\text{ < }}{{\text{c}}^{\text{2}}}$$\Leftrightarrow 5 - 2\sqrt 2 {\text{ < 11 - 6}}\sqrt {\text{2}}$$\Leftrightarrow 4\sqrt 2 {\text{ < 6}} \Leftrightarrow {\left( {{\text{4}}\sqrt {\text{2}} } \right)^2}{\text{ < }}{{\text{6}}^{\text{2}}}$$\Leftrightarrow 32{\text{ < 36}}$(đúng) Vậy phương trình vô nghiệm.