Question

resolvendo a equação 1/(sen x)^2 - 1/(cos x)^2 - 1/(tg x)^2 - 1/(sec x)^2 - 1/(cos x)^2 - 1/(cotg x)^2 = - 3, obtem-se

Answer 1

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Resolver a equação trigonométrica:

$\mathsf{\dfrac{1}{(sen\,x)^2}-\dfrac{1}{(cos\,x)^2}-\dfrac{1}{(tg\,x)^2}-\dfrac{1}{(sec\,x)^2}-\dfrac{1}{(cos\,x)^2}-\dfrac{1}{(cotg\,x)^2}=-3}\\\\\\ \mathsf{\bigg(\dfrac{1}{sen\,x}\bigg)^{\!\!2}-\bigg(\dfrac{1}{cos\,x}\bigg)^{\!\!2}-\bigg(\dfrac{1}{tg\,x}\bigg)^{\!\!2}-\bigg(\dfrac{1}{sec\,x}\bigg)^{\!\!2}-\bigg(\dfrac{1}{cos\,x}\bigg)^{\!\!2}-\bigg(\dfrac{1}{cotg\,x}\bigg)^{\!\!2}=-3}\\\\\\ \mathsf{(cossec\,x)^2-(sec\,x)^2-(cotg\,x)^2-(cos\,x)^2-(sec\,x)^2-(tg\,x)^2=-3}$


Rearrumando a equação,

$\mathsf{\big[(cossec\,x)^2-(cotg\,x)^2\big]-(cos\,x)^2-2(sec\,x)^2-(tg\,x)^2=-3}\\\\ \mathsf{1-(cos\,x)^2-2(sec\,x)^2-(tg\,x)^2=-3}\\\\ \mathsf{1+3=(cos\,x)^2+2(sec\,x)^2+(tg\,x)^2}\\\\ \mathsf{4=(cos\,x)^2+2(sec\,x)^2+(tg\,x)^2}\\\\ \mathsf{4=(cos\,x)^2+\dfrac{2}{(cos\,x)^2}+\bigg(\dfrac{sen\,x}{cos\,x}\bigg)^{\!\!2}}$

$\mathsf{4=(cos\,x)^2+\dfrac{2}{(cos\,x)^2}+\dfrac{(sen\,x)^2}{(cos\,x)^2}}\\\\\\ \mathsf{4=(cos\,x)^2+\dfrac{2}{(cos\,x)^2}+\dfrac{1-(cos\,x)^2}{(cos\,x)^2}}\\\\\\ \mathsf{4=(cos\,x)^2+\dfrac{2+1-(cos\,x)^2}{(cos\,x)^2}}\\\\\\ \mathsf{4=(cos\,x)^2+\dfrac{3-(cos\,x)^2}{(cos\,x)^2}}$


Faça a seguinte mudança de variável:

$\mathsf{(cos\,x)^2=t\qquad\quad(0<t<1)\qquad(i)}$


e a equação fica

$\mathsf{4=t+\dfrac{3-t}{t}}\\\\\\ \mathsf{4-t=\dfrac{3-t}{t}}\\\\\\ \mathsf{t\cdot (4-t)=3-t}\\\\ \mathsf{4t-t^2=3-t}\\\\ \mathsf{0=3-t-4t+t^2}$


$\mathsf{t^2-5t+3=0}\quad\rightarrow\quad\left\{\! \begin{array}{l} \mathsf{a=1}\\\mathsf{b=-5}\\\mathsf{c=3} \end{array} \right.\\\\\\ \mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=(-5)^2-4\cdot 1\cdot 3}\\\\ \mathsf{\Delta=25-12}\\\\ \mathsf{\Delta=13}$

$\mathsf{t=\dfrac{-b\pm \sqrt{\Delta}}{2a}}\\\\ \mathsf{t=\dfrac{-(-5)\pm \sqrt{13}}{2\cdot 1}}\\\\\\ \mathsf{t=\dfrac{5\pm \sqrt{13}}{2}}\\\\\\ \begin{array}{rcl} \mathsf{t=\dfrac{5-\sqrt{13}}{2}}&~\textsf{ ou }~&\mathsf{t=\dfrac{5+\sqrt{13}}{2}} \end{array}$


Atenção, temos de verificar a condição $\mathsf{0<t<1:}$

•   Para $\mathsf{t=\dfrac{5-\sqrt{13}}{2}:}$

Observe que

$\mathsf{0<9<13}\\\\ \mathsf{0<3^2<13}\\\\ \mathsf{0<3<\sqrt{13}}\\\\ \mathsf{0<5-2<\sqrt{13}}\\\\ \mathsf{2<5<2+\sqrt{13}}\\\\ \mathsf{2-\sqrt{13}<5-\sqrt{13}<2}$


Portanto,

$\mathsf{5-\sqrt{13}<2}\\\\ \mathsf{\dfrac{5-\sqrt{13}}{2}<1\qquad\quad(ii)}$


Por outro lado, também temos que

$\mathsf{0<13<25}\\\\ \mathsf{0<13<5^2}\\\\ \mathsf{0<\sqrt{13}<5}\\\\ \mathsf{\sqrt{13}<5}\\\\ \mathsf{0<5-\sqrt{13}}\\\\ \mathsf{0<\dfrac{5-\sqrt{13}}{2}\qquad\quad(iii)}$


Por $\mathsf{\mathsf{(ii)}}$ e $\mathsf{\mathsf{(iii)}},$ conseguimos mostrar que

$\mathsf{0<\dfrac{5-\sqrt{13}}{2}<1\qquad\quad\checkmark}$


•   Para $\mathsf{t=\dfrac{5+\sqrt{13}}{2}:}$

Facilmente verifica-se que

$\mathsf{2<5<5+\sqrt{13}}\\\\ \mathsf{2<5+\sqrt{13}}\\\\ \mathsf{1<\dfrac{5+\sqrt{13}}{2}}\qquad\textsf{(n\~ao serve)}\qquad\diagup\!\!\!\!\!\diagdown$


A única solução para a equação quadrática em $\mathsf{t}$ é

$\mathsf{t=\dfrac{5-\sqrt{13}}{2}}$


Voltando à variável $\mathsf{x}:$

$\begin{array}{l} \mathsf{(cos\,x)^2=\dfrac{5-\sqrt{13}}{2}}\\\\ \mathsf{cos\,x=\pm\,\sqrt{\dfrac{5-\sqrt{13}}{2}}}\\\\ \begin{array}{rcl} \mathsf{cos\,x=-\,\sqrt{\dfrac{5-\sqrt{13}}{2}}}&~\textsf{ ou }~&\mathsf{cos\,x=\sqrt{\dfrac{5-\sqrt{13}}{2}}}\\\\ \mathsf{x=\pm\,arccos\bigg(\!\!-\sqrt{\dfrac{5-\sqrt{13}}{2}}\bigg)+k\cdot 2\pi}&~\textsf{ ou }~&\mathsf{x=\pm\,arccos\bigg(\sqrt{\dfrac{5-\sqrt{13}}{2}}\bigg)+k\cdot 2\pi} \end{array} \end{array}$


onde $\mathsf{k}$ é um inteiro.


Conjunto solução:

$\footnotesize\begin{array}{l} \mathsf{S=\left\{x\in\mathbb{R:}~x=\pm\,arccos\bigg(\!\!-\sqrt{\dfrac{5-\sqrt{13}}{2}}\bigg)+k\cdot 2\pi~~ou~~x=\pm\,arccos\bigg(\sqrt{\dfrac{5-\sqrt{13}}{2}}\bigg)+k\cdot 2\pi,~~k\in\mathbb{Z}\right\}}. \end{array}$


Bons estudos! :-)


Tags:   equação trigonométrica condição de existência identidade mudança de variável segundo grau quadrática álbegra trigonometria