341If you let x^7 = t, then we are looking for a such that
t^2 -at+1 is a factor of t^{22} -at^{11}+1
So every root of t^2-at+1 = 0 should be a root of
t^{22} -at^{11}+1 = 0.
Another way of writing this is that whenever t + \frac{1}{t} = a, we must have t^{11} + \frac{1}{t^{11}} = a
Now the problem mercifully wants us to only find a \in (-2,2).
This is a big hint which actually gives the clue that the equations are to be written in the manner above.
For we if we let a = 2 \cos \theta, (and the domain of a allows such a substitution), then t = \cos \theta + i \sin \theta and its conjugate \bar{t} are the two roots.
So, now the equation simply is 2 \cos 11 \theta =2 \cos \theta
and correspondingly a = 2 \cos \theta for every \theta satisfying the
above equation.