Seja $y = \sin x \cos x$ . Do enunciado, podemos manipular a equação de tal modo que$$(\sin x + \cos x)^2 = (-2\sqrt 2 \sin x \cos x)^2 \implies 1+2y = 8y^2$$$$y^2 - \frac{y}{4} = \frac{1}{8} \implies \left( y - \frac{1}{8}\right)^2 = \frac{1}{8} + \frac{1}{64} = \left(\frac{3}{8}\right)^2$$Dessa forma, temos:$$\left| y - \frac{1}{8}\right| = \frac{3}{8} \implies y = \frac{1 \pm 3}{8} = \left\{-\frac{1}{4},\space \frac{1}{2} \right\}$$ Como $y = \Large{\frac{\sin {2x}}{2}}$, então$$\frac{\sin {2x}}{2} = -\frac{1}{4} \implies \sin {2x} = -\frac{1}{2}$$$$\color{red}{x = \frac{7\pi}{12} + 2k\pi}$$$$\color{red}{x = \frac{11\pi}{12}} + \pi \implies \frac{23\pi}{12} +2k\pi$$$$\frac{\sin {2x}}{2} = \frac{1}{2} \implies \sin {2x} = 1$$$$\color{red}{x = \frac{\pi}{4} + \pi} \implies x = \frac{5\pi}{4} + 2k\pi, \space \space k\in \mathbb{Z}$$Portanto$$\boxed{x = \left\{\frac{5\pi}{4},\space \frac{23\pi}{12},\space \frac{7\pi}{12} \right \} + 2k\pi , \space \space k \in \mathbb{Z}}$$