$ {\sin ^2}x + 2\sin x\cos x - 2{\cos ^2}x = \dfrac{1}{2}\\ \Leftrightarrow {\sin ^2}x + 2\sin x\cos x - 2{\cos ^2}x = \dfrac{1}{2}\left( {{{\cos }^2}x + {{\sin }^2}x} \right)\\ \Leftrightarrow \dfrac{1}{2}{\sin ^2}x + 2\sin x\cos x - \dfrac{5}{2}{\cos ^2}x = 0\\ \Leftrightarrow \left( {\sin x - \cos x} \right)\left( {\sin x + 5\cos x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin x - \cos x = 0\\ \sin x = - 5\cos x \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \sqrt 2 \sin \left( {x - \dfrac{\pi }{4}} \right) = 0\\ \tan x = - 5 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x - \dfrac{\pi }{4} = k\pi \\ x = \arctan \left( { - 5} \right) + k\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{4} + k\pi \\ x = \arctan \left( { - 5} \right) + k\pi \end{array} \right.\\ x \in \left( {0;2\pi } \right) \Rightarrow x = \dfrac{\pi }{4} + k\pi \left( {k \in \mathbb{Z}} \right) \to A $
