[latex]\begin{cases} 2cos^2x+1=2\sqrt{2}sinx \\ sinxcosy=\frac{1}{2} \end{cases}[/latex]

[latex]\begin{cases} 2cos^2x+1=2\sqrt{2}sinx \\ sinxcosy=\frac{1}{2} \end{cases} \\\ sinx=a \\\ siny=b \\\ cos^2x=1-a^2 \begin{cases} 2(1-a^2)+1=2\sqrt{2}a \\ ab=\frac{1}{2} \end{cases} \\\ \begin{cases} 2-2a^2+1=2\sqrt{2}a \\ ab=\frac{1}{2} \end{cases} 2a^2+2\sqrt{2}a-3=0 \\\ D=2+6=8 \\\ a \neq \frac{-\sqrt{2}-2\sqrt{2}}{2} = \frac{-3\sqrt{2}}{2} <-1 \\\ a =\frac{-\sqrt{2}+2\sqrt{2}}{2} = \frac{\sqrt{2}}{2} \\\ sinx= \frac{\sqrt{2}}{2} [/latex] [latex]x=(-1)^k \frac{\pi}{4} +\pi k, k\in Z \\\ \frac{ \sqrt{2} }{2} cosy= \frac{1}{2} \\\ cosy=\frac{ \sqrt{2} }{2} \\\ y=\pm \frac{\pi}{4} +2\pi n , n\in Z[/latex] Ответ: [latex]x=(-1)^k \frac{\pi}{4} +\pi k, k\in Z \\\ y=\pm \frac{\pi}{4} +2\pi n , n\in Z[/latex]