[tex]\displaystyle\bf\\sin^2x+cos^2x=1\\\\sin^2x=1-cos^2x\\\\sinx=\sqrt{1-cos^2x}=\sqrt{1-\left(\frac{3}{5}\right)^\b2}}=\sqrt{1-\frac{9}{25}}=\\\\\\=\sqrt{\frac{25-9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}\\\\\\\boxed{\bf~sinx=\frac{4}{5}}[/tex]
Raspuns corect a)
