Pré-Vestibular ⇒ (PUCCAMP) Função Trigonométrica Tópico resolvido

[FelipeMP]

(PUCCAMP) - Determinar [tex3]0\leq x<2\pi [/tex3] que verifique [tex3]tg\left(x+\frac{\pi }{4}\right)>0[/tex3].

Resposta

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{[tex3]x\in \mathbb{R}|0\leq x<\frac{\pi }{4} [/tex3] ou [tex3]\frac{3\pi }{4}< x < \frac{5\pi }{4}[/tex3] ou [tex3]\frac{7\pi }{4} < x <2\pi [/tex3]}

Grato desde já.

[csmarceloMOD]

É direto que:

[tex3]\tan\theta>0\rightarrow\begin{cases}0<\theta<\frac{\pi}{2}\\\text{ou}\\(0+\pi=\pi)<\theta<\(\frac{\pi}{2}+\pi=\frac{3\pi}{2}\)\end{cases}[/tex3]

No caso,

[tex3]\theta=x+\frac{\pi}{4}[/tex3]

Substituindo,

[tex3]\tan\(x+\frac{\pi}{4}\)>0\rightarrow\begin{cases}0<x+\frac{\pi}{4}<\frac{\pi}{2}\\\text{ou}\\\pi<x+\frac{\pi}{4}<\frac{3\pi}{2}\end{cases}[/tex3]

[tex3]\tan\(x+\frac{\pi}{4}{\color{red}-\frac{\pi}{4}}\)>0\rightarrow\begin{cases}0{\color{red}-\frac{\pi}{4}}<x+\frac{\pi}{4}{\color{red}-\frac{\pi}{4}}<\frac{\pi}{2}{\color{red}-\frac{\pi}{4}}\\\text{ou}\\\pi{\color{red}-\frac{\pi}{4}}<x+\frac{\pi}{4}{\color{red}-\frac{\pi}{4}}<\frac{3\pi}{2}{\color{red}-\frac{\pi}{4}}\end{cases}[/tex3]

[tex3]\tan x>0\rightarrow\begin{cases}-\frac{\pi}{4}<x<\frac{\pi}{4}\rightarrow\begin{cases}0\leq x<\frac{\pi}{4}\\\text{ou}\\\frac{7\pi}{4}<x<2\pi\end{cases}\\\text{ou}\\\frac{3\pi}{4}<x<\frac{5\pi}{4}\end{cases}[/tex3]

[FelipeMP]

csmarcelo, muito obrigado! Eu usei soma de tangentes, e me compliquei todo rs.