Start by making a graph of the problem, taking into account that x is in the fourth quadrant and the triangle formed has a hypotenuse of 2 and an adjacent side of √2
use the Pythagorean theorem to find the missing side
![\begin{gathered} a^2+b^2=c^2 \\ a=\sqrt[]{c^2-b^2} \\ a=\sqrt[]{(2)^2-(\sqrt[]{2})^2} \\ a=\sqrt[]{4-2} \\ a=\sqrt[]{2} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/9swwucnigaw4u9rljticahr46mujpz7k8a.png)
then,
find the value of six(X)
![\begin{gathered} \sin x=(op)/(hy) \\ \sin x=-\frac{\sqrt[]{2}}{2} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/eb7pduhshrmrshgeuz1oegxf0dq7132u0z.png)
use the double angle identities for cos
![\begin{gathered} \cos 2x=\cos ^2x-\sin ^2x \\ \cos 2x=(\frac{\sqrt[]{2}}{2})^2-(-\frac{\sqrt[]{2}}{2})^2 \\ \cos 2x=(2)/(4)-(2)/(4) \\ \cos 2x=(1)/(2)-(1)/(2) \\ \cos 2x=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/yush4hrc7pp6na2cikbzc9jsi17gc9owmg.png)