Dissertativa 10 - 2ª Fase - Dia 1 - ITA 2027

Substituindo $\cos \left(2\mathrm{x}\right)=2{\cos }^2\mathrm{x}-1$ e $\cos \left(2\mathrm{x}\right)=1-2{\mathrm{sen}}^2\mathrm{x}$:

$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{2{\cos }^2\mathrm{x}}+\sqrt{2{\mathrm{sen}}^2\mathrm{x}}-\mathrm{sen} \mathrm{x}-\cos \mathrm{x}$

$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{2}\left|\cos \mathrm{x}\right|+\sqrt{2}\left|\mathrm{sen} \mathrm{x}\right|-\mathrm{sen} \mathrm{x}-\cos \mathrm{x}$

Caso 1: solução no primeiro quadrante.

Tem-se $\mathrm{sen} \mathrm{x} >0$ e $\cos \mathrm{x}>0$, logo:

$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{2}\cos \mathrm{x}+\sqrt{2}\mathrm{sen} \mathrm{x}-\mathrm{sen} \mathrm{x}-\cos \mathrm{x}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\left(\sqrt{2}-1\right)\left(\mathrm{sen} \mathrm{x} + \cos \mathrm{x}\right)$

Fazendo $\mathrm{sen} \mathrm{x} + \cos \mathrm{x}=\sqrt{2} \mathrm{sen}\left(\mathrm{x}+\frac{\mathrm{\pi }}{4}\right)$:

$\mathrm{f}\left(\mathrm{x}\right)=\left(2-\sqrt{2}\right)\mathrm{sen}\left(\mathrm{x}+\frac{\mathrm{\pi }}{4}\right)$

Como $0+2k\pi \le \mathrm{x}\le \frac{\mathrm{\pi }}{2}+2k\pi$, tem-se $\frac{\mathrm{\pi }}{4}+2k\pi \le \mathrm{x}+\frac{\mathrm{\pi }}{4}\le \frac{3\mathrm{\pi }}{4}+2k\pi$.

Portanto, $\frac{\sqrt{2}}{2}\le \mathrm{sen}\left(\mathrm{x}+\frac{\mathrm{\pi }}{4}\right)\le 1$.

Assim:

$\frac{\sqrt{2}}{2}<\frac{\mathrm{f}\left(\mathrm{x}\right)}{2-\sqrt{2}}\le 1\Rightarrow \sqrt{2}-1\le \mathrm{f}\left(\mathrm{x}\right)\le 2-\sqrt{2}\Rightarrow 0,4\le \mathrm{f}\left(\mathrm{x}\right)\le 0,6$

Não há inteiro nesse intervalo.

Caso 2: solução no terceiro quadrante.

Tem-se $\mathrm{sen} \mathrm{x}<0$ e $\cos \mathrm{x}<0$, logo:

$\mathrm{f}\left(\mathrm{x}\right)=-\sqrt{2}\cos \mathrm{x}-\sqrt{2}\mathrm{sen} \mathrm{x}-\mathrm{sen} \mathrm{x} - \cos \mathrm{x} \Rightarrow \mathrm{f}\left(\mathrm{x}\right)=-\left(\sqrt{2}+1\right)\left(\mathrm{sen} \mathrm{x} + \cos \mathrm{x}\right)$

Novamente, faz-se $\mathrm{sen} \mathrm{x}+\cos \mathrm{x}=\sqrt{2}\mathrm{sen}\left(\mathrm{x}+\frac{\mathrm{\pi }}{4}\right)$:

$\mathrm{f}\left(\mathrm{x}\right)=-\left(2+\sqrt{2}\right)\mathrm{sen}\left(\mathrm{x}+\frac{\mathrm{\pi }}{4}\right)$

Como $\mathrm{\pi }+2\mathrm{k\pi }\le \mathrm{x}\le \frac{3\mathrm{\pi }}{2}+2k\pi$, tem-se $\frac{5\mathrm{\pi }}{4}+2k\pi \le \mathrm{x}+\frac{\mathrm{\pi }}{4}\le \frac{7\mathrm{\pi }}{4}+2k\pi$.

Portanto, $-1\le \mathrm{sen} \left(\mathrm{x}+\frac{\mathrm{\pi }}{4}\right)\le -\frac{\sqrt{2}}{2}$. 

Assim:

$-1\le -\frac{\mathrm{f}\left(\mathrm{x}\right)}{2+\sqrt{2}}\le -\frac{\sqrt{2}}{2}\Rightarrow \sqrt{2}+1\le \mathrm{f}\left(\mathrm{x}\right)\le 2+\sqrt{2}\Rightarrow 2,4\le \mathrm{f}\left(\mathrm{x}\right)\le 3,4$

A única solução inteira ocorre para $\mathrm{f}\left(\mathrm{x}\right)=3$.

Portanto, só há solução para $\boxed{\mathrm{k}=3}$.