For this we use the trigonometric identities
[tex]\begin{gathered} \cos ^2(x)+\sin ^2(x)=1 \\ \tan (x)=\frac{\sin (x)}{\cos (x)} \\ \cot (x)=\frac{\cos (x)}{\sin (x)} \end{gathered}[/tex]Since we have the value of sin (x) we start from the first identity to obtain cos (x):
[tex]\begin{gathered} \sin (x)=\frac{1}{4}\Rightarrow\sin ^2(x)=\frac{1}{4^2}\Rightarrow\sin ^2(x)=\frac{1}{16} \\ \text{Then,} \\ \cos ^2(x)+\sin ^2(x)=1 \\ \cos ^2(x)+\frac{1}{16}=1 \\ \cos ^2(x)=1-\frac{1}{16} \\ \cos ^2(x)=\frac{15}{16} \\ \cos (x)=\pm\sqrt[]{\frac{15}{16}} \\ \text{ Since tangent is negative and by the second propery, we take} \\ \cos (x)=-\sqrt[]{\frac{15}{16}} \\ \cos (x)=\frac{-\sqrt[]{15}}{\sqrt[]{16}} \\ \cos (x)=\frac{-\sqrt[]{15}}{4} \end{gathered}[/tex]Finally, we use the third property to obtain cot (x):
[tex]\begin{gathered} \cot (x)=\frac{\cos(x)}{\sin(x)} \\ \cot (x)=\frac{\frac{-\sqrt[]{15}}{4}}{\frac{1}{4}} \end{gathered}[/tex]Applying the Sandwich Law for Fractions, we have
[tex]\begin{gathered} \cot (x)=\frac{-\sqrt[]{15}\cdot4}{4\cdot1} \\ \cot (x)=-\sqrt[]{15} \end{gathered}[/tex]