Resposta :
Oi, Bianca.
[tex]\sin x=0,6=\frac6{10}=\frac35\\\\ \sin^2x+\cos^2x=1 \Rightarrow \frac9{25}+\cos^2x=1 \Rightarrow \cos^2x=1-\frac9{25} \Rightarrow \\\\ \cos x=\sqrt{\frac{25-9}{25}}=\sqrt{\frac{16}{25}}=\frac45\\\\ \tan x=\frac{\sin x}{\cos x}=\frac35\cdot\frac54 \Rightarrow \boxed{\tan x=\frac34}[/tex]
Observe que:
[tex](\text{sen}~\text{x})^2+(\text{cos}~\text{x})^2=1[/tex]
Como [tex]\text{sen}~\text{x}=0,6[/tex], temos que:
[tex](0,6)^2+(\text{cos}~\text{x})^2=1[/tex]
Logo:
[tex]\text{cos}~\text{x}=\sqrt{1-\dfrac{36}{100}}=\sqrt{\dfrac{64}{100}}=\dfrac{8}{10}=0,8[/tex]
Depois disso, note que:
[tex]\text{tg}~\text{x}=\dfrac{\text{sen}~\text{x}}{\text{cos}~\text{x}}=\dfrac{0,6}{0,8}=\dfrac{\frac{6}{10}}{\frac{8}{10}}=\dfrac{6}{8}=\dfrac{3}{4}[/tex]