Questão 10 - 1ª Fase - ITA 2026

Considere a segunda equação do sistema:

$\begin{array}{ccc}\frac{\mathrm{sen}\left(\mathrm{y}\right)}{\cos \left(\mathrm{x}\right)}+\frac{\cos \left(\mathrm{x}\right)}{\mathrm{sen}\left(\mathrm{y}\right)}& =& 2\end{array}$

Seja $\begin{array}{ccc}\frac{\mathrm{sen}\left(\mathrm{y}\right)}{\cos \left(\mathrm{x}\right)}& =& \mathrm{a}.\end{array}$Assim:

$\mathrm{a}+\frac{1}{\mathrm{a}}=2\to \mathrm{a}=1$

Então:

$\mathrm{sen}\left(\mathrm{y}\right)=\cos \left(\mathrm{x}\right)$

Como $\mathrm{x},\mathrm{y} \in ]0, \mathrm{\pi }/2[ : \mathrm{x}=\frac{\mathrm{\pi }}{2}-\mathrm{y}$.

Desse modo, a primeira equação do sistema pode ser reescrita como:

$\tan \left(\mathrm{y}\right)·\tan \left(\frac{\mathrm{y}}{2}\right)=\cos \left(2·\left(\frac{\mathrm{\pi }}{2}-\mathrm{y}\right)\right)·\sec \left(\mathrm{y}\right)$

Assim:

$\frac{\mathrm{sen}\left(\mathrm{y}\right)}{\cos \left(\mathrm{y}\right)}·\tan \left(\frac{\mathrm{y}}{2}\right)=\cos \left(\mathrm{\pi }-2\mathrm{y}\right)·\frac{1}{\cos \left(\mathrm{y}\right)}$

Simplificando a expressão, tem-se:

$\mathrm{sen}\left(\mathrm{y}\right)·\tan \left(\frac{\mathrm{y}}{2}\right)=-\cos \left(2\mathrm{y}\right)$

Utilizando a identidade $\cos \left(2\mathrm{\alpha }\right)=1-2{\mathrm{sen}}^2\mathrm{\alpha }:$

$\mathrm{sen}\left(\mathrm{y}\right)·\tan \left(\frac{\mathrm{y}}{2}\right)=2{\mathrm{sen}}^2\left(\mathrm{y}\right)-1$

Utilizando a identidade: $\mathrm{sen}\left(\mathrm{\alpha }\right)=\frac{2\tan \left({\displaystyle \frac{\alpha }{2}}\right)}{1+{\tan }^2\left({\displaystyle \frac{\alpha }{2}}\right)}:$

$\frac{2{\tan }^2\left({\displaystyle \frac{y}{2}}\right)}{1+{\tan }^2\left({\displaystyle \frac{y}{2}}\right)}=2{\left(\frac{2\tan \left({\displaystyle \frac{y}{2}}\right)}{1+{\tan }^2\left({\displaystyle \frac{y}{2}}\right)}\right)}^2-1$

Fazendo ${\tan }^2\left(\frac{y}{2}\right)=\mathrm{k}:$

$\frac{2\mathrm{k}}{1+\mathrm{k}}=2{\left(\frac{2\sqrt{\mathrm{k}}}{1+\mathrm{k}}\right)}^2-1$

Resolvendo a equação:

$\mathrm{k}=\frac{1}{3} \mathrm{ou} \mathrm{k}=1$

Considerando que $\mathrm{y}\in ]0, \mathrm{\pi }/2[, 0<\tan \left(\frac{\mathrm{y}}{2}\right)<1\to 0<\mathrm{k}<1.$ Então:

${\tan }^2\left(\frac{\mathrm{y}}{2}\right)=\frac{1}{3}\to \tan \left(\frac{\mathrm{y}}{2}\right)=\frac{\sqrt{3}}{3}\to \frac{\mathrm{y}}{2}=\frac{\mathrm{\pi }}{6}\to \mathrm{y}=\frac{\mathrm{\pi }}{3}$

Como $\mathrm{x}=\frac{\mathrm{\pi }}{2}-\mathrm{y}=\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{3}=\frac{\mathrm{\pi }}{6}$,

$\mathrm{x}·\mathrm{y}=\frac{\mathrm{\pi }}{6}·\frac{\mathrm{\pi }}{3}=\boxed{\frac{{\mathrm{\pi }}^2}{18}}$