Respostas

2016-10-24T07:39:55-02:00
\large\begin{array}{l} \mathsf{tg\,x=-\,\dfrac{3}{5}}\\\\ \mathsf{\dfrac{sen\,x}{cos\,x}=-\,\dfrac{3}{5}}\\\\ \mathsf{5\,sen\,x=-3\,cos\,x} \end{array}


\large\begin{array}{l} \textsf{Elevando os dois lados ao quadrado, obtemos}\\\\ \mathsf{(5\,sen\,x)^2=(-3\,cos\,x)^2}\\\\ \mathsf{25\,sen^2\,x=9\,cos^2\,x\qquad}\textsf{(mas }\mathsf{sen^2\,x=1-cos^2\,x}\textsf{)}\\\\ \mathsf{25\cdot (1-cos^2\,x)=9\,cos^2\,x}\\\\ \mathsf{25-25\,cos^2\,x=9\,cos^2\,x}\\\\ \mathsf{25=9\,cos^2 x+25\,cos^2\,x}\\\\ \mathsf{25=34\,cos^2\,x}\\\\ \mathsf{cos^2\,x=\dfrac{25}{34}} \end{array}

\large\begin{array}{l} \mathsf{cos\,x=\pm \sqrt{\dfrac{25}{34}}}\\\\ \mathsf{cos\,x=\pm\,\dfrac{5}{\sqrt{34}}\qquad}\left(\textsf{mas }\mathsf{cos\,x=\dfrac{1}{sec\,x}}\right)\\\\ \mathsf{\dfrac{1}{sec\,x}=\pm\,\dfrac{5}{\sqrt{34}}}\\\\ \mathsf{sec\,x=\pm\,\dfrac{\sqrt{34}}{5}}\\\\ \end{array}


\large\begin{array}{l} \textsf{Como a tangente \'e negativa, x s\'o pode ser do segundo}\\\textsf{ou do quarto quadrantes.}\\\\ \bullet~~\textsf{Se x for do segundo quadrante, ent\~ao}\\\\ \mathsf{sec\,x<-1}\\\\ \boxed{\begin{array}{c}\mathsf{sec\,x=-\,\dfrac{\sqrt{34}}{5}} \end{array}}\\\\\\ \bullet~~\textsf{Se x for do quarto quadrante, ent\~ao}\\\\ \mathsf{sec\,x<1}\\\\ \boxed{\begin{array}{c}\mathsf{sec\,x=\dfrac{\sqrt{34}}{5}} \end{array}} \end{array}


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\large\textsf{Bons estudos! :-)}