If $\cos x+2\cos 3x=\cos y$ and $\sin x+2\sin 3x=\sin y$ then $\cos 2x=$

Squaring the given relations, we get

${\cos }^{2}x+4{\cos }^{2}3x+4\cos x\cos 3x={\cos }^{2}y$ 

$$\begin{gathered} and {\sin }^{2}x+4{\sin }^{2}3x+4\sin x\sin 3x={\sin }^{2}y \\ Adding the above relations, we get \\ 1+4+4\cos (3x-x)=1 \end{gathered}$$

$\Rightarrow 4\cos 2x=-4$

   $\Rightarrow \cos 2x=-1$