Resposta:
Explicação passo a passo:
sen²x + cos²x = 1
[tex](-\frac{1}{3})^2+cos^2x=1\\\\cos^2x=1-\frac{1}{9} \\\\cos^2x=\frac{8}{9}\\\\cos^2x=\frac{2\sqrt{2} }{3} \\\\cosx=-\frac{2\sqrt{2} }{3} ,~~~pois~~x~~pertence~~ao~~terceiro~~quadrante.~~cosx<0\\\\tgx=-\frac{1}{3} .(-\frac{3}{2\sqrt{3} } )=\frac{1}{2\sqrt{3} } =\frac{\sqrt{3} }{6}[/tex]
[tex]cos2a=cos^2a-sen^2a=cos^2a-(1-cosa) = 2cos^2a-1=2(\frac{2}{3})^2-1= 2.\frac{4}{9}-1=\frac{8}{9}-1 =\frac{-1}{9}[/tex]
