cos(4v) + cos(v) = 0 I am given the following equation: 
$$\cos 4v + \cos v = 0$$
My attempt: 
$$\cos4v = -\cos v$$
$$\cos4v = \cos(\pi \pm v)$$
$$4v = \pm \pi \pm v + 2\pi n$$

$$4v_1 = \pi + v_1 + 2\pi n$$
$$4v_2 = -\pi - v_2 + 2\pi n$$
$$4v_3 = +\pi - v_3 + 2\pi n$$
$$4v_4 = -\pi + v_4 + 2\pi n$$

$$v_1 = \frac{\pi}{3} + \frac{2\pi n}{3}$$
$$v_2 = -\frac{\pi}{5} + \frac{2\pi n}{5}$$
$$v_3 = \frac{\pi}{5} + \frac{2\pi n}{5}$$
$$v_4 = -\frac{\pi}{3} + \frac{2\pi n}{3}$$
However, the answer is simply the positive solutions i.e: 
$$v_1 = \frac{\pi}{3} + \frac{2\pi n}{3}$$
$$v_3 = \frac{\pi}{5} + \frac{2\pi n}{5}$$
Why? What am I missing?

Using Chebyshev polynomials of the first kind,
$$ T_4(x)+T_1(x) = 8x^4-8x^2+x+1 = (x+1)(8x^2(x-1)+1) = (1+x)(1-2x)(1+2x-4x^2) $$
hence the solutions are given by $\cos(v)\in\left\{-1,\frac{1}{2},\frac{1-\sqrt{5}}{4},\frac{1+\sqrt{5}}{4}\right\}$ or:
$$ v\in \left\{\pm\frac{\pi}{5},\pm\frac{\pi}{3},\pm\frac{3\pi}{5},\pm\pi\right\}+2\pi\mathbb{Z}.$$
