Resposta :

MATHSPHIS
[tex]sen^2x+cos^2x=1 \\ \\ cos^2x=1-sen^2x \\ \\ cos^x=\sqrt{1-sen^2x} \\ \\ cos x=\sqrt{1-\left(\frac{3}{4}\right)^2} \\ \\ cos x=\sqrt{1-\frac{9}{16}} \\ \\ cos x =\sqrt{\frac{7}{16}} \\ \\ \boxed{cosx=\frac{\sqrt7}{4}}[/tex]

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