Given sine of x equals negative 12 over 13 and cos x > 0, what is the exact solution of cos 2x?

Use the definition of sine to find the known sides of the unit circle right triangle.
[tex]\sin x=\frac{opposite}{hypotenuse}=-\frac{12}{13}[/tex]Find the adjacent side of the unit circle triangle. Since the hypotenuse and opposite sides are known, use the Pythagorean theorem to find the remaining side.
[tex]\text{Adjacent =}\sqrt[]{hypotenuse^2-opposite^2}[/tex]Replace the known values in the equation.
[tex]Adjacent=\sqrt[]{13^2-(-12)^2}=\sqrt[]{169-144}=\sqrt[]{25}=5[/tex]Find the value of cosine.
[tex]\begin{gathered} \cos x=\frac{adjacent}{hypotenuse} \\ \cos x=\frac{5}{13} \end{gathered}[/tex]Next, we have
[tex]\cos 2x=\cos ^2x-\sin ^2x[/tex]Then, replace the values of sin(x) and cos(x)
[tex]\cos 2x=(\frac{5}{13})^2-(-\frac{12}{13})^2=\frac{25}{169}-\frac{144}{169}=\frac{25-144}{169}=-\frac{119}{169}[/tex]Answer:
[tex]-\frac{119}{169}[/tex]