ANSWER:
[tex]\begin{gathered} \sin 2x=\frac{12}{13} \\ \cos 2x=-\frac{5}{13} \\ \tan 2x=-\frac{12}{5} \end{gathered}[/tex]STEP-BY-STEP EXPLANATION:
We know that cotagent is given as follows:
[tex]\begin{gathered} \cot x=\frac{\text{ adjacent }}{\text{ opposite}} \\ \text{ therefore} \\ \text{adjacent = 2} \\ \text{ oppoiste = 3} \\ \text{ hypotenuse =}\sqrt[]{2^2+3^2}=\sqrt[]{4+9}=\sqrt[]{13} \end{gathered}[/tex]Therefore:
sin 2x:
[tex]\begin{gathered} \sin 2x=2\sin x\cdot\cos x \\ \sin x=\frac{\text{ opposite}}{\text{ hypotenuse}}=\frac{3}{\sqrt[]{13}} \\ \cos x=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{2}{\sqrt[]{13}} \\ \sin 2x=2\sin x\cdot\cos x=2\cdot\frac{3}{\sqrt{13}}\cdot\frac{2}{\sqrt{13}}=\frac{12}{13} \end{gathered}[/tex]cos 2x:
[tex]\begin{gathered} \cos 2x=\cos ^2x-\sin ^2x \\ \sin ^2x=\mleft(\frac{3}{\sqrt{13}}\mright)^2=\frac{9}{13} \\ \cos ^2x=\mleft(\frac{2}{\sqrt{13}}\mright)^2=\frac{4}{13} \\ \cos 2x=\frac{4}{13}-\frac{9}{13}=-\frac{5}{13} \end{gathered}[/tex]tan 2x:
[tex]\tan 2x=\frac{\sin 2x}{\cos 2x}=\frac{\frac{12}{13}}{-\frac{5}{13}}=-\frac{12}{5}[/tex]