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

$\Rightarrow 2\left[\cos x+\cos 3x\right]=\cos y$

$\Rightarrow 2.2\cos 2x.\cos x=\cos y$

$\Rightarrow 4\cos 2x.\cos x=\cos y\to \left(1\right)$

$\Rightarrow 2\sin x+2\sin 3x=\sin y$

$\Rightarrow 2\left(\sin x+\sin 3x\right)=\sin y$

$\Rightarrow 4\sin 2x.\cos x=\sin y\to \left(2\right)$

$\left(1\right)÷\left(2\right)$

$\cot 2x=\cot y\Rightarrow 2x=y\Rightarrow \cos 2x=\cos y$

$\Rightarrow \cos 2x=\cos y=2\cos x+2\cos 3x$ $=4\cos 2x.\cos x$

$\Rightarrow \cos 2x\left(1-4\cos x\right)=0$

$\Rightarrow 1-4\cos x=0\Rightarrow \cos x=\frac{1}{4}$

$\Rightarrow \cos 2x=2{\cos }^{2}x-1$

$=2\left(\frac{1}{16}\right)-1$

$=\frac{1}{8}-1=-\frac{7}{8}$