cos(ax) = cos((a-1)x + x)
expand and continue in that fashion
So you have:
2cos5x + 2cos4x + 2cos3x + 2cos2x + 2cosx + 1 = 0
And I said:
cos(ax) = cos((a-1)x + x)
expand and continue in that fashion
So, for instance, lets look at the case where a = 5:
cos5x = cos(4x + x)
Now that is of the form cos(A+B), which we know how to expand:
cos(A+B) = cosA.cosB - sinA.sinB
So, for cos5x:
cos5x = cos(4x + x)
. = cos(4x).cos(x) - sin(4x).sin(x)
And then you can do the same sort of expansions with sin(4x), cos(4x), cos(3x) etc., until you are satisfied you have simplified it enough to solve.