Question

se x=8pi/3 radianos, o valor da expressão sen x - cos x/ sen x + cos x é um número real pertencente a:
a)[4; infinito [
b) [3;4]
c) [1/2;3]
d) ]-raiz de 3; 1/2[
e)]- infinito; - raiz de 3[

URGENTE!!!!!!!!!

Answer 1

Resposta:

$\boxed{\mathtt{B}}$

Explicação passo-a-passo:

$\\ \displaystyle \mathsf{x = \frac{8\pi}{3} = \frac{6\pi + 2\pi}{3} = \frac{6\pi}{3} + \frac{2\pi}{3} = 2\pi + \frac{2\pi}{3}} \\\\ \mathsf{\Rightarrow \boxed{\mathsf{x = \frac{2\pi}{3}}}}$

Assim, temos que:

$\\ \displaystyle \mathsf{\frac{\sin x - \cos x}{\sin x + \cos x} =} \\\\\\ \mathsf{\frac{\sin 120^o - \cos 120^o}{\sin 120^o + \cos 120^o} =} \\\\\\ \mathsf{\frac{\sin 60^o - (- \cos 60^o)}{\sin 60^o + (- \cos 60^o)} =} \\\\\\ \mathsf{\frac{\frac{\sqrt{3}}{2} + \frac{1}{2}}{\frac{\sqrt{3}}{2} - \frac{1}{2}} =} \\\\\\ \mathsf{\frac{\sqrt{3} + 1}{2} \cdot \frac{2}{\sqrt{3} - 1} =}$

$\\ \displaystyle \mathsf{\frac{\sqrt{3} + 1}{\sqrt{3} - 1} \cdot \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)} =} \\\\\\ \mathsf{\frac{(\sqrt{3} + 1)^2}{3 - 1} =} \\\\\\ \mathsf{\frac{3 + 2\sqrt{3} + 1}{2} =} \\\\\\ \mathsf{\frac{4 + 2\sqrt{3}}{2} =} \\\\ \boxed{\boxed{\mathsf{2 + \sqrt{3}}}}$