Respostas

2013-08-05T18:26:24-03:00
\sin^{2}x+\cos^{2}=1\\\\
(\dfrac{3}{5})^2+\cos^{2}=1\\\\
\dfrac{9}{25}+\cos^{2}=1\\\\
\cos^{2}=1-\dfrac{9}{25}=\dfrac{25-9}{25}\\\\
\cos^{2}=\dfrac{16}{25}\\\\
\cos x=\pm\sqrt{\dfrac{16}{25}}\\\\
\cos x=\pm\dfrac{4}{5}}

Como 90^{\circ}, o cosseno é negativo, então:

\cos x=-\dfrac{4}{5}}
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