Question

$\begin{array}{l}\mathsf{Calcular~sen~x~~e~~cos~x~~sabendo~que~5\cdot sec~x-3\cdot tg^2~x=1}\end{array}$


*Trigonometria, relações fundamentais.*

Answer 1

$5sec(x)-3tg^2(x) = 1\\ \\ Lembrando\ que:\\ \\ sec(x)=\frac{1}{cos(x)}\ e\ tg(x)=\frac{sen(x)}{cos(x)}\\ \\ \frac{5}{cos(x)} - 3(\frac{sen^2(x)}{cos^2(x)})=1\\ \\ \frac{5cos(x)-3sen^2(x)}{cos^2(x)}=1\\ \\ 5cos(x)-3sen^2(x) = cos^2(x)\\ \\ 5cos(x)-3(1-cos^2(x)) = cos^2(x)\\ \\ 5cos(x)-3+3cos^2(x) = cos^2(x)\\ \\ 2cos^2(x)+5cos(x)-3=0\\ \\ \boxed{cos(x) = y}\\ \\ 2y^2+5y-3=0\\ \\ \Delta=5^2-4(2)(-3)=25+24 = 49\\ \\ y' = \frac{-5+7}{4} = \frac{2}{4} = \frac{1}{2}\\ \\ y''=\frac{-5-7}{4} = \frac{-12}{4} = -3$

$cos(x)=y\\ \\ \boxed{cos(x) = \frac{1}{2}}\\ \\ cos(x)=y\\ \\ cos(x) = -3\\ \\ \boxed{Como\ -1 \leq cos(x) \leq 1, esse\ valor\ n\~ao\ entra\ na\ resposta.}\\ \\ Logo:\\ \\ sen^2(x)+cos^2(x)=1\\ \\ sen^2(x) +(\frac{1}{2})^2 = 1\\ \\ sen^2(x) = 1-\frac{1}{4}\\ \\ sen^2(x) = \frac{3}{4}\\ \\ sen(x) = \pm \sqrt{\frac{3}{4}} \\ \\ \large\boxed{sen(x)=\pm\frac{\sqrt{3}}{2}}}$