Đáp án:
$\begin{array}{l}
{\cos ^2}x - \sin 2x = \sqrt 2 {\cos ^2}\left( {\frac{\pi }{2} + x} \right)\\
\Rightarrow {\cos ^2}x - 2.\sin x.\cos x = \sqrt 2 {\sin ^2}x\\
\Rightarrow {\cos ^2}x - 2\sin x.\cos x - \sqrt 2 {\sin ^2}x = 0\\
+ Khi:\cos x = 0\\
\Rightarrow \sin x = 0\left( {ktm} \right)\\
+ Khi:\cos x \ne 0\\
\Rightarrow 1 - 2\frac{{\sin x}}{{\cos x}} - \sqrt 2 .\frac{{{{\sin }^2}x}}{{{{\cos }^2}x}} = 0\\
\Rightarrow 1 - 2.\tan x - \sqrt 2 .{\tan ^2}x = 0\\
\Rightarrow \left[ \begin{array}{l}
\tan x = 0,39\\
\tan x = - 1,8
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = {21^0} + k{.180^0}\\
x = {61^0} + k{.180^0}
\end{array} \right.\\
Do:x \in \left( {0;{{360}^0}} \right)\\
\Rightarrow \left[ \begin{array}{l}
x = {21^0}\\
x = {201^0}\\
x = {61^0}\\
x = {241^0}
\end{array} \right.
\end{array}$