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 $$2(\cos\angle A+1)(\cos\angle B+1)(\cos\angle C+1)=\cos(\angle A-\angle B)+\cos(\angle B-\angle C)+\cos(\angle C-\angle A)+\cos\angle A+\cos\angle B+\cos\angle C+2$$
 A: Re-writing the equation as
$$0 = 2\prod (\cos A + 1) - \sum \cos(B-C) - \sum \cos A - 2$$
we begin by multiplying-out the product, and carrying-on from there:
$$\begin{align}
0 &= 2\cos A \cos B \cos C + 2\sum \cos B \cos C \color{blue}{+ 2\sum \cos A} \color{red}{+ 2} \\[4pt]
&\quad-\sum\cos(B-C) \color{blue}{- \sum\cos A} \color{red}{- 2} \\[8pt]
&= 2\cos A \cos B \cos C + \sum\left(\; 2 \cos B \cos C + \cos A -\cos(B-C)\;\right) \\[8pt]
&= 2\cos A \cos B \cos C + \sum\left(\; 2 \cos B \cos C - \cos(B+C) -\cos(B-C)\;\right) \\[8pt]
&= 2\cos A \cos B \cos C + \sum\left(\; 2 \cos B \cos C - 2 \cos B \cos C \;\right)\\[8pt]
&= 2 \cos A \cos B \cos C
\end{align}$$
Thus, the exercise reduces to 

$$B = 2 C \qquad\text{and}\qquad \cos A \cos B \cos C = 0$$

which is readily solved.
Edit. Since the solutions were posted as comments, I'll provide arguments for them.
Note that the equation $\cos A \cos B \cos C = 0$ implies that one of $A$, $B$, $C$ is a right angle. Taking these case by case ...

$$\begin{align}
A = 90^\circ &\quad\implies\quad 90^\circ = B+C = 2 C + C = 3C \implies \color{blue}{( A, B, C ) = (90^\circ, 60^\circ, 30^\circ)} \\[4pt]
B = 90^\circ &\quad\implies\quad C = B/2 = 45^\circ \implies \color{blue}{(A, B, C) = (45^\circ, 90^\circ, 45^\circ)} \\[4pt]
C = 90^\circ &\quad\implies\quad B = 2 C = 180^\circ \implies \color{red}{\text{invalid triangle}}
\end{align}$$

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