$\alpha + \beta + \gamma = \pi$ , show that $\cos 2\alpha + \cos 2\beta + \cos 2\gamma + 2\cos\alpha \cos\beta \cos\gamma = 1$ 
$\cos 2\alpha + \cos 2\beta + \cos 2\gamma + 2\cos\alpha \cos\beta \cos\gamma = 1$

I really didn't know how to solve this problem and I am very unused to the utilization of trigonometric identities, I was wondering if I may have some assistance in this problem with detailed explanations
I was thinking more along the lines of making all of the angles in terms of one angle
 A: It looks like you got a wrong identity. 
Similar to/Extend Mehta's answer, use the identities:
$$\begin{aligned}\cos(2\alpha)+\cos(2\beta) &= 2\cos(\alpha+\beta)\cos(\alpha-\beta),\\
&= -2\cos(\gamma)\cos(\alpha-\beta)
\end{aligned}$$
and
$$\cos(2\gamma)\color{\red}+1 = 2 \cos^2\gamma.$$
Adding them up we obtain
$$\begin{aligned}
\cos(2\alpha)+\cos(2\beta)+\cos(2\gamma) +1 &= -2\cos\gamma\cos(\alpha-\beta)+2\cos^2\gamma\\
&= -2\cos\gamma(\cos(\alpha-\beta)+\cos(\alpha+\beta))\\
&= -4\cos\gamma\cos\alpha\cos\beta.
\end{aligned}$$
So the identity we have here is
$$\cos(2\alpha)+\cos(2\beta)+\cos(2\gamma)+1 = - 4 \cos\alpha\cos\beta\cos\gamma,\tag{1}$$
which is different from what you ask for. We can check by specific values of $(\alpha,\beta,\gamma)$. For example $(\pi/2,\pi/4,\pi/4)$ turns (1) into $0=0$ while your identity would be $-1=1$.
A: I think you mean the following problem.

Let $\alpha+\beta+\gamma=\pi$. Prove that:
  $$\cos^2\alpha+\cos^2\beta+\cos^2\gamma+2\cos\alpha\cos\beta\cos\gamma=1$$

We need to prove that
$$\cos^2\alpha+\cos^2\beta-2\cos\alpha\cos\beta\cos(\alpha+\beta)=\sin^2(\alpha+\beta)$$ or
$$\cos^2\alpha+\cos^2\beta-2\cos\alpha\cos\beta(\cos\alpha\cos\beta-\sin\alpha\sin\beta)=(\sin\alpha\cos\beta+\cos\alpha\sin\beta)^2$$ or
$$\cos^2\alpha+\cos^2\beta-2\cos^2\alpha\cos^2\beta=\sin^2\alpha\cos^2\beta+\cos^2\alpha\sin^2\beta$$ or
$$\cos^2\alpha(1-\sin^2\beta)+\cos^2\beta(1-\sin^2\alpha)-2\cos^2\alpha\cos^2\beta=0,$$
which is obvious.
Done!
A: Your first thought is good, but it's not possible to write all the angles in terms of just one angle.  Instead, we can express $\gamma = \pi - \alpha - \beta$, which might help!
$$\begin{align}&\cos (2\alpha) + \cos (2 \beta) + \cos (2\pi - 2\alpha - 2\beta) + 2 \cos (\alpha) \cos (\beta) \cos (\pi - \alpha - \beta) \\=& \cos (2\alpha) + \cos (2 \beta) + \cos (2\alpha + 2\beta) - 2 \cos (\alpha) \cos (\beta) \cos ( \alpha +\beta)
\end{align}$$
Then, expand out using standard identities to finish off.
