[tex]\displaystyle\\\\\text{Consider ca ai vrut sa scrii: }~x\in \Big(0,~\frac{\pi}{2}\Big)\\\\\text{Folosim formula: }~\sin^2x+\cos^2x=1~~\Longleftrightarrow~~\sin x=\sqrt{1-\cos^2x}\\ \\\text{Rezolvare:}\\\\\cos x=\frac{3}{5}\\\\\sin x=\sqrt{1-\cos^2x}=\sqrt{1-\Big(\frac{3}{5} \Big)^2}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{25-9}{25}}=\sqrt{\frac{16}{25}}=\boxed{\bf\frac{4}{5}}[/tex]