Estude! Com Prof. Caju

[tex3]h'(x)=sen\ x+cos\ x\\\\h'(x)\geq 0\rightarrow sen\ x+cos\ x\geq 0\rightarrow \\\\\rightarrow cos\ x+cos\ \left(\frac{\pi}{2}-x\right)\geq 0\rightarrow \\\\\rightarrow 2cos\ \left(\frac{x+\frac{\pi}{2}-x}{2}\right)cos\ \left(\frac{x-\frac{\pi}{2}+x}{2}\right)\geq 0\rightarrow \\\\\rightarrow cos\ \left(x-\frac{\pi}{4}\right)\geq 0\rightarrow \begin{cases}
\frac{\pi}{4}+2k\pi\leq x\leq \frac{3\pi}{4}+2k\pi \\
\frac{7\pi}{4}+2k\pi\leq x\leq \frac{9\pi}{4}+2k\pi
\end{cases}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]h'(x)=sen\ x+cos\ x\\\\h'(x)\geq 0\rightarrow sen\ x+cos\ x\geq 0\rightarrow \\\\\rightarrow cos\ x+cos\ \left(\frac{\pi}{2}-x\right)\geq 0\rightarrow \\\\\rightarrow 2cos\ \left(\frac{x+\frac{\pi}{2}-x}{2}\right)cos\ \left(\frac{x-\frac{\pi}{2}+x}{2}\right)\geq 0\rightarrow \\\\\rightarrow cos\ \left(x-\frac{\pi}{4}\right)\geq 0\rightarrow \begin{cases}
\frac{\pi}{4}+2k\pi\leq x\leq \frac{3\pi}{4}+2k\pi \\
\frac{7\pi}{4}+2k\pi\leq x\leq \frac{9\pi}{4}+2k\pi
\end{cases}[/tex3]

Por que que a solução: [tex3]\frac{7\pi}{4}+2k\pi\leq x\leq \frac{9\pi}{4}+2k\pi[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{7\pi}{4}+2k\pi\leq x\leq \frac{9\pi}{4}+2k\pi[/tex3]

foi descartada?

Gabarito: [tex3]\frac{\pi}{4}+2k\pi\leq x\leq \frac{3\pi}{4}+2k\pi[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{\pi}{4}+2k\pi\leq x\leq \frac{3\pi}{4}+2k\pi[/tex3]