F6 Trigo Equation Problem

1 + cosA + cos2A + ... + cosnA = 0

sin[(n + 1)A/2]cos(nA / 2) / sin(A / 2) = 0

As sin(A / 2) =/= 0, we have A =/= 0 or 2π

Now, sin[(n + 1)A/2]cos(nA / 2) = 0

sin[(n + 1)A/2] = 0 or cos(nA / 2) = 0

(n + 1)A/2 = 0 or π or 2π or nA / 2 = π/2 or 3π/2

A = 0 (rej.), 2π/(n + 1) or 4π/(n + 1) or A = π/n or 3π/n

But note that the range of A is [0 , 2π]

Therefore, the general solution of A:

A = 2π/(n + 1), 4π/(n + 1), π/n or 3π/m

where n is any positive integers and m is any positive integers >= 2.

(Note that m =/= 1.)