Questão 7 - 1ª Fase - ITA 2027

Considera-se a identidade:

$\mathrm{tg}\left(\frac{\mathrm{x}}{2}\right)=\frac{\mathrm{senx}}{1+\mathrm{cosx}}$

Aplicada à equação dada, tem-se:

$$\begin{gathered} \frac{{\mathrm{sen}}^2\mathrm{x}}{1-\mathrm{cosx}}·\frac{\mathrm{senx}}{1+\mathrm{cosx}}=1-\mathrm{senx}-\mathrm{cosx} \\ \frac{{\mathrm{sen}}^3\mathrm{x}}{1-{\cos }^2\mathrm{x}}=1-\mathrm{senx}-\mathrm{cosx} \\ \frac{{\mathrm{sen}}^3\mathrm{x}}{{\mathrm{sen}}^2\mathrm{x}}=1-\mathrm{senx}-\mathrm{cosx} \\ \mathrm{senx}=1-\mathrm{senx}-\mathrm{cosx} \\ 2\mathrm{senx}=1-\mathrm{cosx} \\ 2\mathrm{senx}+\mathrm{cosx}=1 \end{gathered}$$

Utilizando as identidades:

$\mathrm{senx}=\frac{2\mathrm{tg}{\displaystyle \frac{\mathrm{x}}{2}}}{1+{\mathrm{tg}}^2{\displaystyle \frac{\mathrm{x}}{2}}}$ e $\mathrm{cosx}=\frac{1-{\mathrm{tg}}^2{\displaystyle \frac{\mathrm{x}}{2}}}{1+{\mathrm{tg}}^2{\displaystyle \frac{\mathrm{x}}{2}}}$

Tem-se:

$$\begin{gathered} 2·\frac{2\mathrm{tg}{\displaystyle \frac{\mathrm{x}}{2}}}{1+{\mathrm{tg}}^2{\displaystyle \frac{\mathrm{x}}{2}}}+\frac{1-{\mathrm{tg}}^2{\displaystyle \frac{\mathrm{x}}{2}}}{1+{\mathrm{tg}}^2{\displaystyle \frac{\mathrm{x}}{2}}}=1 \\ 4·\mathrm{tg}\frac{\mathrm{x}}{2}+1-{\mathrm{tg}}^2\frac{\mathrm{x}}{2}=1+{\mathrm{tg}}^2\frac{\mathrm{x}}{2} \\ 2·{\mathrm{tg}}^2\frac{\mathrm{x}}{2}-4\mathrm{tg}\frac{\mathrm{x}}{2}=0 \\ \mathrm{tg}\frac{\mathrm{x}}{2}\left(\mathrm{tg}\frac{\mathrm{x}}{2}-2\right)=0 \end{gathered}$$

Caso 1: $\mathrm{tg}\frac{\mathrm{x}}{2}-2=0\Rightarrow \mathrm{tg}\frac{\mathrm{x}}{2}=2$

$\frac{\mathrm{x}}{2}=\mathrm{arctg}\left(2\right)+\mathrm{k\pi }\Rightarrow \mathrm{x}=2\mathrm{arctg}\left(2\right)+2\mathrm{k\pi }$

Caso 2: $\mathrm{tg}\frac{\mathrm{x}}{2}=0\Rightarrow \mathrm{senx}=0$ e $\mathrm{cosx}=1$ (absurdo, pois fere a condição de existência da equação inicial)

Assim, a única solução é:

$\boxed{\mathrm{x}=2\mathrm{arctg}\left(2\right)+2\mathrm{k\pi }}$

A seguir, apresenta-se uma segunda opção de solução para esse item.

Sabe-se que:

$\left\{\begin{array}{l}\mathrm{sen} \mathrm{x}=2\left(\mathrm{sen} \frac{\mathrm{x}}{2}\right)\left(\cos \frac{\mathrm{x}}{2}\right)\\ 1-\cos \frac{\mathrm{x}}{2}=2{\mathrm{sen}}^2\frac{\mathrm{x}}{2}\\ \mathrm{tg}\left(\frac{\mathrm{x}}{2}\right)=\frac{\mathrm{sen} \frac{\mathrm{x}}{2}}{\cos \frac{\mathrm{x}}{2}}\end{array}\right.$

Voltando para a expressão original, tem-se:

$$\begin{gathered} \frac{4{\mathrm{sen}}^2\frac{\mathrm{x}}{2}{\cos }^2\frac{\mathrm{x}}{2}}{2{\mathrm{sen}}^2\frac{\mathrm{x}}{2}}·\frac{\mathrm{sen} \frac{\mathrm{x}}{2}}{\cos \frac{\mathrm{x}}{2}}=2{\mathrm{sen}}^2\frac{\mathrm{x}}{2}-2\mathrm{sen} \frac{\mathrm{x}}{2}\cos \frac{\mathrm{x}}{2} \\ 2\mathrm{sen} \frac{\mathrm{x}}{2}\cos \frac{\mathrm{x}}{2}=2\mathrm{sen} \frac{\mathrm{x}}{2}\left(\mathrm{sen} \frac{\mathrm{x}}{2}-\cos \frac{\mathrm{x}}{2}\right) \end{gathered}$$

Como $1-\mathrm{cosx}=2{\mathrm{sen}}^2\frac{\mathrm{x}}{2}\ne 0 \Rightarrow \mathrm{sen}\frac{\mathrm{x}}{2}\ne 0$

Ou seja:

$$\begin{gathered} \cos \frac{\mathrm{x}}{2}=\mathrm{sen} \frac{\mathrm{x}}{2}-\cos \frac{\mathrm{x}}{2} \\ \mathrm{tg} \frac{\mathrm{x}}{2}=2 \\ \frac{\mathrm{x}}{2}=\mathrm{arctg}\left(2\right)+\mathrm{k\pi } \\ \boxed{\mathrm{x}=2\mathrm{arctg}\left(2\right)+2\mathrm{k\pi }} \end{gathered}$$