Respuesta :

[tex]\cos{x}=- \frac{2}{ \sqrt{13} } \\ \\ \cos^2{x}=(- \frac{2}{ \sqrt{13} } )^2= \frac{4}{13} \\ \\ \sin^2{x}=1-\cos^2{x}=1-\frac{4}{13} =\frac{9}{13} \\ \\ \sin{x}= \sqrt{\frac{9}{13} } = \frac{3}{ \sqrt{13} } \\ \\ \sin{2x}=2\sin{x}\cos{x}=2\frac{3}{ \sqrt{13} }(- \frac{2}{ \sqrt{13} } )=- \frac{12}{13} \\ \\ \cos{2x}=\cos^2{x}-\sin^2{x}=\frac{4}{13} -\frac{9}{13} =-\frac{5}{13} \\ \\ \tan{2x}= \frac{\sin{2x}}{\cos2x}}= \frac{- \frac{12}{13} }{-\frac{5}{13} } =\frac{12}{5} [/tex]

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