Question

Se sen x 2/3 o valor da tan x é

Answer 1

Supondo x do primeiro quadrante:

$cos(x)=\sqrt{1-sen^2(x)}\\ cos(x)=\sqrt{1-(\frac{2}{3})^2}\\ cos(x)=\sqrt{1-\frac{4}{9}}\\ cos(x)=\sqrt{\frac{5}{9}}\\ \boxed{cos(x)=\frac{\sqrt5}{3}\\} \\ tg(x)=\frac{sen(x)}{cos(x)}\\ tg(x)=\frac{\frac{2}{3}}{\frac{\sqrt5}{3}}\\ \\ \boxed{tg(x)=\frac{2}{\sqrt5}=\frac{2\sqrt5}{5}}$

Answer 2

Explicação passo-a-passo:

Pela relação fundamental da trigonometria:

$\sf sen^2~x+cos^2~x=1$

$\sf \Big(\dfrac{2}{3}\Big)^2+cos^2~x=1$

$\sf cos^2~x+\dfrac{4}{9}=1$

$\sf cos^2~x=1-\dfrac{4}{9}$

$\sf cos^2~x=\dfrac{9-4}{9}$

$\sf cos^2~x=\dfrac{5}{9}$

Assim:

$\sf tg~x=\dfrac{sen~x}{cos~x}$

$\sf tg^2~x=\dfrac{sen^2~x}{cos^2~x}$

$\sf tg^2~x=\dfrac{\frac{4}{9}}{\frac{5}{9}}$

$\sf tg^2~x=\dfrac{4}{9}\cdot\dfrac{9}{5}$

$\sf tg^2~x=\dfrac{36}{45}$

$\sf tg^2~x=\dfrac{4}{5}$

$\sf tg~x=\pm\sqrt{\dfrac{4}{5}}$

$\sf tg~x=\pm\dfrac{2}{\sqrt{5}}$

$\sf tg~x=\pm\dfrac{2}{\sqrt{5}}\cdot\dfrac{\sqrt{5}}{\sqrt{5}}$

$\sf tg~x=\pm\dfrac{2\sqrt{5}}{5}$

$\sf tg~x=\dfrac{2\sqrt{5}}{5}~ou~tg~x=\dfrac{-2\sqrt{5}}{5}$