Question

Dou 30 pontos. Ajuda ai

Answer 1

Questão 06.

$\mathrm{sen\,}\alpha=\dfrac{1}{3}$


Elevando os dois lados ao quadrado, temos

$\mathrm{sen^2\,}\alpha=\left(\dfrac{1}{3} \right )^{\!\!2}\\\\\\ \mathrm{sen^2\,}\alpha=\dfrac{1}{9}~~~~~~~(\text{mas }\mathrm{sen^2\,}\alpha=1-\cos^2 \alpha)\\\\\\ 1-\cos^2 \alpha=\dfrac{1}{9}\\\\\\ 1-\dfrac{1}{9}=\cos^2\alpha\\\\\\ \dfrac{9}{9}-\dfrac{1}{9}=\cos^2\alpha\\\\\\ \dfrac{9-1}{9}=\cos^2\alpha\\\\\\ \cos^2\alpha=\dfrac{8}{9}$

$\cos \alpha=\pm \sqrt{\dfrac{8}{9}}\\\\\\ \cos \alpha=\pm \dfrac{\sqrt{8}}{\sqrt{9}}\\\\\\ \cos \alpha=\pm \dfrac{2\sqrt{2}}{3}$


Como $\alpha$ é um arco do 2º quadrante, o seu cosseno é negativo. Logo,

$\boxed{\begin{array}{c}\cos \alpha=-\dfrac{2\sqrt{2}}{3} \end{array}}$

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Questão 07.

$\mathrm{sen\,}x=3\cos x$


Elevando os dois lados ao quadrado, temos

$\mathrm{sen^2\,}x=(3\cos x)^2\\\\ \mathrm{sen^2\,}x=9\cos^2 x~~~~~~~(\text{mas }\mathrm{sen^2\,}x=1-\cos^2 x)\\\\ 1-\cos^2 x=9\cos^2 x\\\\ 1=9\cos^2 x+\cos^2 x\\\\ 1=10\cos^2 x\\\\ \cos^2 x=\dfrac{1}{10}\\\\\\ \cos x=\pm\sqrt{\dfrac{1}{10}}\\\\\\ \cos x=\pm \dfrac{1}{\sqrt{10}}$


Como $x$ é do 3º quadrante, o seu cosseno é negativo. Portanto,

$\boxed{\begin{array}{c}\cos x=-\,\dfrac{1}{\sqrt{10}} \end{array}}$


Encontrando o seno de $x:$

$\mathrm{sen\,}x=3\cos x\\\\\\ \mathrm{sen\,}x=3\cdot\left(-\,\dfrac{1}{\sqrt{10}} \right )\\\\\\ \boxed{\begin{array}{c}\mathrm{sen\,}x=-\,\dfrac{3}{\sqrt{10}} \end{array}}$