قرار می‌دهیم $\operatorname{Sin}x=\operatorname{Cos}x=y$، بنابراین:  ${{(\operatorname{Sin}x+\operatorname{Cos}x)}^{2}}={{y}^{2}}\Rightarrow {{\operatorname{Sin}}^{2}}x+{{\operatorname{Cos}}^{2}}x+2\operatorname{Sin}x\operatorname{Cos}x={{y}^{2}}\Rightarrow 1+\operatorname{Sin}2x={{y}^{2}}\Rightarrow \operatorname{Sin}2x={{y}^{2}}-1$   بنابراین: $3\sqrt{2}(\operatorname{Sin}x+\operatorname{Cos}x)\operatorname{Sin}2x+5=0\Rightarrow 3\sqrt{2}y+({{y}^{2}}-1)+5=0\Rightarrow {{y}^{2}}+3\sqrt{2}y+4=0,\Delta =18-16=2\Rightarrow y=\frac{-3\sqrt{2}\pm \sqrt{2}}{2}\Rightarrow y=-\sqrt{2}*y=-2\sqrt{2}$  پس $\operatorname{Sin}x+\operatorname{Cos}x=-\sqrt{2}$ یا $\operatorname{Sin}x+\operatorname{Cos}x=-2\sqrt{2}$، واضح است که حاصل $\operatorname{Sin}x+\operatorname{Cos}x$ نمی‌تواند برابر $-2\sqrt{2}$ باشد (چرا؟). بنابراین: $\operatorname{Sin}x+\operatorname{Cos}x=-\sqrt{2}\Rightarrow \sqrt{2}\operatorname{Sin}(x+\frac{\pi }{4})=-\sqrt{2}\Rightarrow \operatorname{Sin}(x+\frac{\pi }{4})=-1\to x=\frac{\pi }{4}=2k\pi -\frac{\pi }{2}\Rightarrow x=2k\pi -\frac{3\pi }{2}$