The number of ordered pairs $(\alpha, \beta)$ , where $\alpha, \beta \in (-\pi, \pi)$ satisfying $\cos(\alpha - \beta)=1$... 
The number of ordered pairs $(\alpha, \beta)$ , where $\alpha, \beta \in (-\pi, \pi)$ satisfying $\cos(\alpha - \beta)=1 $ and $\cos(\alpha + \beta) = \frac{1}{e}$ is...

What I have tried:
$$\cos(\alpha - \beta)=1 \implies \alpha - \beta = 0$$
$$\cos(\alpha + \beta) = \frac{1}{e} \implies \cos^{-1} \frac{1}{e}$$
From here onwards I do not know what else to do. Please Help.
 A: Hint The condition $\cos \theta = 1$ holds iff $\theta \in 2 \pi \Bbb Z$. So, since the open interval $(-\pi, \pi)$ has length $2 \pi$, we have $\alpha = \beta$. The remaining condition thus simplifies to $\cos 2 \alpha = \frac{1}{e}$.
Now, use the fact that $\cos \tau$ achieves each value in $(-1, 1)$ exactly twice per period in $\tau$.
A: We have $\alpha=\beta$ so $\cos2\alpha=\frac1e$. Since $0<\frac1e<1$, there are four possible values of $2\alpha=\cos^{-1}\frac1e$ (since $2\alpha\in(-2\pi,2\pi)$), and the same number of solutions $(\alpha,\beta)$.
A: Given that $$\cos(\alpha-\beta)=1$$
$$\cos(\alpha-\beta)=\cos0$$
$$\alpha-\beta=0$$
$$\alpha=\beta$$
Now, we have $\cos(\alpha+\beta)=\dfrac1e$
Note that we have $\alpha=\beta$, so
$$cos(\alpha+\alpha)=\dfrac1e$$
$$\cos2\alpha=\dfrac1e$$
Since, the angle is between $[\pi,\pi]$
$$-\pi\le\alpha\le\pi$$
$$-2\pi\le2\alpha\le2\pi$$
So, we have angles from $-2\pi$ to $0$ and from $0$ to $2\pi$.
So, the value of $\cos$ is positive and is in the quadrants $I,IV$.
So, there are $4$ pairs of $(\alpha,\beta)$ in total. 
