Question

Determine os valores de θ ∈ ] 0 , 2π [ tais que

$log_{tg(\theta)} \ e^{sen(\theta)} \ \geq \ 0$

Para esclarecer a base do logaritmo é a $tg(\theta)$ e o logaritmando é $e^{sen(\theta)}$

Answer 1

$\mathsf{\ell og_~{tg\theta }}\ \mathsf{e^{sen\theta}\geq 0}$


Primeiro vamos a condição de existência


$\begin{matrix}\textsf{Tg deve existir}\\\\\mathsf{\theta\neq \dfrac{\pi}{2}}\\\\\mathsf{\theta\neq \dfrac{3\pi}{2}}\end{matrix}\begin{vmatrix}\mathsf{tg\theta\neq 1}\\\\\mathsf{\theta\neq \dfrac{\pi}{4}}\\\\\mathsf{\theta\neq \dfrac{5\pi}{4}}\end{vmatrix}\begin{matrix}\mathsf{tg\theta\neq -1}\\\\\mathsf{\theta\neq \dfrac{3\pi}{4}}\\\\\mathsf{\theta\neq \dfrac{7\pi}{4}}\end{matrix}\begin{vmatrix}\mathsf{tg\ \theta>0}\\\\\mathsf{0<\theta<\dfrac{\pi}{2}}\quad\textsf{ou}\quad\\\\\mathsf{\pi<\theta<\dfrac{3\pi}{2}}\end{vmatrix}\begin{matrix}\mathsf{e^{sen\theta}\geq 0}\\\\\mathsf{\theta\in\mathbb{R}}\\\end{matrix}$



Agora vamos a resolução


$\texttt{Se}$    $\mathtt{tg\theta\ \textgreater \ 1}$    $\texttt{ou seja,se}$    $\mathsf{\frac{\pi}{4}\ \textless \ \theta\ \textless \ \frac{\pi}{2}}$    $\texttt{ou}$    $\mathsf{\frac{5\pi}{4}\ \textless \ \theta\ \textless \ \frac{3\pi}{2}}$:


$\mathsf{e^{sen\theta} \geq tg\theta\ ^{0}}\\\\ \mathsf{e^{sen\theta} \geq 1}\\\\ \mathsf{e^{sen\theta} \geq e\ ^{0}}\\\\ \mathsf{sen\theta \geq 0}\\\\ \mathsf{0 \geq \theta \geq \pi}$


$\mathsf{S'=}\left ( \mathsf{\dfrac{\pi}{4},\dfrac{\pi}{2}} \right )$



$\texttt{Se}$    $\mathtt{tg\theta\ \textless \ 1}$    $\texttt{ou seja,se}$    $\mathsf{0\ \textless \ \theta\ \textless \ \frac{\pi}{4}}$    $\texttt{ou}$    $\mathsf{\pi\ \textless \ \theta\ \textless \ \frac{5\pi}{4}}$:


$\mathsf{e^{sen\theta} \leq tg\theta\ ^{0}}\\\\ \mathsf{e^{sen\theta} \leq 1}\\\\ \mathsf{e^{sen\theta} \leq e\ ^{0}}\\\\ \mathsf{sen\theta \leq 0}\\\\ \mathsf{\pi \leq \theta \leq 2\pi}$


$\mathsf{S''=}\left ( \mathsf{\pi,\dfrac{5\pi}{4}} \right )$



Resposta final: S' ∪ S"

$\boxed{\mathsf{S=}\left ( \mathsf{\dfrac{\pi}{4},\dfrac{\pi}{2}} \right )\cup \left ( \mathsf{\pi,\dfrac{5\pi}{4}} \right )}}$


Bons estudos! =)