Question

URGENTE:
Existem exatamente 6 valores reais de x, com 0 ⩽ x < 2π, para os quais se verifica a igualdade tgˆ3x = tg x. Qual é a soma desses valores?

A- π
B- 2π
C- 3π
D- 4π
E- 5π

Answer 1

Resposta:

Letra D.

Explicação passo-a-passo:

$\tan^3{x}=\tan{x}\ \therefore\ \boxed{\tan{x}(\tan^2{x}-1)=0}$

Caso 1:

$\tan{x}=0\ \therefore\ x=\arctan{0}\ \therefore\ \boxed{x=n\pi, n\in\mathbb{Z}}\\\\ 0\leq x<2\pi\ \therefore\ \boxed{x=\left\{0\right\}}$

Caso 2:

$\tan^2{x}-1=0\ \therefore\ \tan{x}=\pm1$

Caso 2.1:

$\tan{x}=-1\ \therefore\ x=\arctan{(-1)}\ \therefore\ \boxed{x=(4n+3)\dfrac{\pi}{4},n\in\mathbb{Z}}$

$0\leq x<2\pi\ \therefore\ \boxed{x=\dfrac{3\pi}{4}}\ \ \text{ou}\ \ \boxed{x=\dfrac{7\pi}{4}}$

Caso 2.2:

$\tan{x}=1\ \therefore\ x=\arctan{1}\ \therefore\ \boxed{x=(4n+1)\dfrac{\pi}{4},n\in\mathbb{Z}}$

$0\leq x<2\pi\ \therefore\ \boxed{x=\dfrac{\pi}{4}}\ \ \text{ou}\ \ \boxed{x=\dfrac{5\pi}{4}}$

Conjunto solução:

$\boxed{x=\left\{0,\dfrac{\pi}{4},\dfrac{3\pi}{4},\dfrac{5\pi}{4},\dfrac{6\pi}{4} \right\}}$

Dessa forma, a soma dos valores será:

$0+\dfrac{\pi}{4}+\dfrac{3\pi}{4}+\dfrac{5\pi}{4}+\dfrac{7\pi}{4}=\dfrac{16\pi}{4}=\boxed{\boxed{4\pi}}$