Respuesta :

[tex]\bf cos(x)+\sqrt{2}=-cos(x)\implies 2cos(x)+\sqrt{2}=0\implies 2cos(x)=-\sqrt{2} \\\\\\ cos(x)=-\cfrac{\sqrt{2}}{2}\implies \measuredangle x= \begin{cases} \frac{3\pi }{4}\\ \frac{5\pi }{4} \end{cases}[/tex]

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