Respuesta :

[tex]\begin{gathered} 2\sin \mleft(2x\mright)+\sqrt{2}=0 \\ 2\sin \mleft(2x\mright)=-\sqrt{2} \\ \sin \mleft(2x\mright)=-\frac{\sqrt{2}}{2} \\ \text{General solution for }\sin \mleft(2x\mright)=-\frac{\sqrt{2}}{2} \\ 2x=\frac{5\pi}{4}+2\pi n,\: 2x=\frac{7\pi}{4}+2\pi n \\ x=\frac{5\pi}{8}+\pi n,\: x=\frac{7\pi}{8}+\pi n \end{gathered}[/tex]

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