Question

Resolva a equação sec²(x) + tg(x) = 1 para 0 ≤ x < 2π.

Answer 1

Resposta:

Explicação passo a passo:

Resolver a equação trigonométrica

    $\mathsf{sec^2(x)+tg(x)=1}$

para $\mathsf{x\in[0,\,2\pi).}$

Aplicando a identidade trigonométrica

    sec²(x) = 1 + tg²(x)

a equação fica

    $\mathsf{\Longleftrightarrow\quad \big(\diagup\!\!\!\! 1+tg^2(x)\big)+tg(x)=\diagup\!\!\!\! 1}\\\\ \mathsf{\Longleftrightarrow\quad tg^2(x)+tg(x)=0}\\\\ \mathsf{\Longleftrightarrow\quad tg(x)\cdot \big(tg(x)+1\big)=0}\\\\ \mathsf{\Longleftrightarrow\quad tg(x)=0\quad ou\quad tg(x)+1=0}\\\\ \mathsf{\Longleftrightarrow\quad tg(x)=0\quad ou\quad tg(x)=-1}\\\\ \mathsf{\Longrightarrow\quad x\in\{0,\,\pi\}\quad ou\quad x\in\left\{\dfrac{3\pi}{4},\,\dfrac{7\pi}{4}\right\}}$

Portanto, o conjunto solução é

    $\mathsf{S=\left\{0,\,\dfrac{3\pi}{4},\,\pi,\,\dfrac{7\pi}{4}\right\}\quad\longleftarrow\quad resposta.}$

Dúvidas? Comente.

Bons estudos! :-)

Answer 2

$\mathtt{{sec}^{2}(x)+tg(x)=1}\\\mathtt{1+{tg}^{2}(x)+tg(x)=1}\\\mathtt{{tg}^{2}(x)+tg(x)=1-1}\\\mathtt{{tg}^{2}(x)+tg(x)=0}$

$\mathtt{tg(x)(tg(x)+1)=0}\\\mathtt{tg(x)=0\to\,x=0~ou~x=\pi}\\\mathtt{tg(x)+1=0}\\\mathtt{tg(x)=-1}$

$\mathtt{x=\dfrac{3\pi}{4}~ou~x=\dfrac{7\pi}{4}}$

$\mathtt{s=\{0,\pi,\dfrac{3\pi}{4},\dfrac{7\pi}{4}\}}$