According to the problem, we have
[tex]\sin \theta=\frac{3}{5}[/tex]Then, we use the Pythagorean Trigonometric Identity
[tex]\begin{gathered} (\sin \theta)^2+(\cos \theta)^2=1 \\ (\frac{3}{5})^2+(\cos \theta)^2=1 \\ \frac{9}{25}+(\cos \theta)^2=1 \\ (\cos \theta)^2=1-\frac{9}{25}=\frac{25-9}{25}=\frac{16}{25} \\ \cos \theta=\frac{4}{5} \end{gathered}[/tex]Now, we use the Double Angle Identity
[tex]\begin{gathered} \cos ^{}2\theta=\cos ^2\theta-\sin ^2\theta \\ \cos 2\theta=\frac{16}{25}-\frac{9}{25}=\frac{16-9}{25}=\frac{7}{25} \end{gathered}[/tex]