$\begin{align}  & 2{{\cos }^{2}}x+\underbrace{2\sin x\cos x}_{\sin 2x}=1\Rightarrow \underbrace{2{{\cos }^{2}}x-1}_{\cos 2x}+\sin 2x=0 \\  & \Rightarrow -\sin 2x=\cos 2x\Rightarrow \sin (-2x)=\cos 2x \\  & \Rightarrow \cos (\frac{\pi }{2}+2x)=\cos 2x \\  & \Rightarrow \frac{\pi }{2}+2x=2k\pi -2x\Rightarrow 4x=2k\pi -\frac{\pi }{2} \\  & \Rightarrow x=\frac{k\pi }{2}-\frac{\pi }{8} \\ \end{align}$