Question

Resolva a equação trigonométrica:

Answer 1

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$\boxed{\begin{array}{c}\sf a~func_{\!\!,}\tilde ao~tangente~\acute e\\\sf negativa~no~2^{\underline o}~e~4^{\underline o}~quadrantes.\\\sf resta~saber~qual~\acute e~o~arco~cuja~tangente~vale~\dfrac{\sqrt{3}}{3}.\\\sf a~resposta~\acute e~\dfrac{\pi}{6}.\\\sf vamos~portanto~achar~os~arcos~c\hat ongruos~a~\dfrac{\pi}{6}\\\sf nos~2^{\underline o}~e~4^{\underline o}~quadrantes.\end{array}}\\\underline{\sf no~2^{\underline o}~quadrante:}\\\sf \pi-\dfrac{\pi}{6}=\dfrac{6\pi-\pi}{6}=\dfrac{5\pi}{6}$

$\underline{\sf no~4^{\underline o}~quadrante:}\\\sf 2\pi-\dfrac{\pi}{6}=\dfrac{12\pi-\pi}{6}=\dfrac{11\pi}{6}\\\sf DA\acute I:\\\sf tg(5x-\pi)=-\dfrac{\sqrt{3}}{3}\\\sf tg(5x-\pi)=tg\left(\dfrac{5\pi}{6}\right)\\\sf 5x-\pi=\dfrac{5\pi}{6}\\\sf 30x-6\pi=5\pi\\\sf 30x=6\pi+5\pi\\\sf 30x=11\pi\\\huge\boxed{\boxed{\boxed{\boxed{\sf x=\dfrac{11\pi}{30}\checkmark+k\pi,k\in\mathbb{Z}}}}}$

$\sf tg(5x-\pi)=tg\left(\dfrac{11\pi}{6}\right)\\\sf 5x-\pi=\dfrac{11\pi}{6}\\\sf30x-6\pi=11\pi\\\sf 30x=11\pi+6\pi\\\sf30x=17\pi\\\huge\boxed{\boxed{\boxed{\boxed{\sf x=\dfrac{17\pi}{30}+k\pi,k\in\mathbb{Z}}}}}$$\sf s=\left\{x\in\mathbb{R}/ x=\dfrac{11\pi}{30}+k\pi,k\in\mathbb{Z}, x=\dfrac{17\pi}{30}+k\pi,k\in\mathbb{Z}\right\}$