Finding a triangle ABC if $2\prod (\cos \angle A+1)=\sum \cos(\angle A-\angle B)+\sum \cos \angle A+2$ Find $\triangle ABC$ if $\angle B=2\angle C$ and $$2(\cos\angle A+1)(\cos\angle B+1)(\cos\angle C+1)=\cos(\angle A-\angle B)+\cos(\angle B-\angle C)+\cos(\angle C-\angle A)+\cos\angle A+\cos\angle B+\cos\angle C+2$$

Hint:$$A+B+C=\pi,B=2C\Rightarrow A=\pi-3C$$
$\cos A=-\cos 3C$
$\cos B=\cos 2C$
$\cos(A-B)=-\cos 5C$
$\cos(B-C)=\cos C$
$\cos(C-A)=-\cos 4C$
