Question

Obtenha a sen x e cos x, sabendo que a raiz quadrada de 3 vale 1,7 e que:

Answer 1

$\boxed{\begin{array}{l}\sf 5\cdot cossec(x)-3\cdot cotg^2(x)=1\\\sf\dfrac{5}{sen(x)}-\dfrac{3cos^2(x)}{sen^2(x)}=1\times(sen^2(x))\\\sf 5\cdot sen(x)-3cos^2(x)=sen^2(x)\\\sf sen^2(x)+3cos^2(x)-5\cdot sen(x)=0\\\sf cos^2(x)=1-sen^2(x)\\\sf sen^2(x)+3\cdot(1-sen^2(x))-5\cdot sen(x)=0\\\sf sen^2(x)+3-3sen^2(x)-5\cdot sen(x)=0\\\sf -2sen^2(x)-5sen(x)+3=0\cdot (-1)\\\sf 2sen^2(x)+5sen(x)-3=0\end{array}}$

$\boxed{\begin{array}{l}\sf\Delta=b^2-4ac\\\sf\Delta=5^2-4\cdot2\cdot(-3)\\\sf\Delta=25+24\\\sf\Delta=49\\\sf sen(x)=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf sen(x)=\dfrac{-5\pm\sqrt{49}}{2\cdot 2}\\\\\sf sen(x)=\dfrac{-5\pm7}{4}\begin{cases}\sf sen(x)=\dfrac{-5+7}{4}=\dfrac{2}{4}=\dfrac{1}{2}\\\\\sf sen (x)=\dfrac{-5-7}{4}=-\dfrac{12}{4}=-3\end{cases}\end{array}}$

$\boxed{\begin{array}{l}\sf sen(x)=\dfrac{1}{2}\\\sf x=\dfrac{\pi}{6}+2k\pi,k\in\mathbb{Z}\\\\\sf x=\dfrac{5\pi}{6}+2k\pi, k\in\mathbb{Z}\\\\\sf S=\bigg\{\dfrac{\pi}{6}+2k\pi,k\in\mathbb{Z}~ou~ \dfrac{5\pi}{6}+2k\pi,k\in\mathbb{Z}\bigg\}\\\sf sen(x)=-3\\\sf S=\bigg\{\bigg\}~pois~-1\leqslant sen(x)\leqslant 1\end{array}}$

$\boxed{\begin{array}{l}\sf cos(x)=cos\bigg(\dfrac{\pi}{6}\bigg)=\dfrac{\sqrt{3}}{2}=\dfrac{1,7}{2}=0,85\end{array}}$