Question

ME AJUDA PFVVVVVVV SZ


Se 0º < x < 90º e tg x = $\sqrt[3]{3}$ , então o valor de $\frac{ sen^{3}x+ cos^{3}x }{ sen^{3}x- cos^{3} x }$ é:

a) 1/2
b) 1
c) 2
d) 5/2
e) 3

Answer 1

$\mathrm{\tan{x}=\sqrt[3]{3}\ \to\ \dfrac{\sin{x}}{\cos{x}}=\sqrt[3]{3}\ \to\ \sin{x}=\cos{x}.\sqrt[3]{3}}\\\\\\ \mathrm{\dfrac{\sin^3{x}+\cos^3{x}}{\sin^3{x}-\cos^3{x}}=\dfrac{(\cos{x}.\sqrt[3]{3})^3+\cos^3{x}}{(\cos{x}.\sqrt[3]{3})^3-\cos^3{x}}=}\\\\\\ \mathrm{=\dfrac{3\cos^3{x}+\cos^3{x}}{3\cos^3{x}-\cos^3{x}}=\dfrac{4\cos^3{x}}{2\cos^3{x}}=\dfrac{4}{2}=\mathbf{2}}$