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?
 A: $n$ is not necessarily positive.
$$ -\dfrac{\pi}{3} = \dfrac{\pi}{3} + \dfrac{2\pi(-1)}{3}$$
etc.
A: Your answers are actually the same as the official answers. If you draw a cast diagram you can see this
A: 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:
$$ v\in \left\{\pm\frac{\pi}{5},\pm\frac{\pi}{3},\pm\frac{3\pi}{5},\pm\pi\right\}+2\pi\mathbb{Z}.$$
A: You can use a cosine identity to simplify:
$$\cos(u)+\cos(w)=2\cos\left(\frac{u+w}{2}\right)\cos\left(\frac{u-w}{2}\right)$$
You then take $u=4v$ and $w=v$:
$$\cos(4v)+\cos(v)=2\cos\left(\frac{4v+v}{2}\right)\cos\left(\frac{4v-v}{2}\right)=0$$
You should then be able to simplify and solve.
A: HINT: use that $$\cos(x)-\cos(y)=-2\sin\left(\frac{x-y}{2}\right)\sin\left(\frac{x+y}{2}\right)$$
sorry i misreaded the post use that
$$\cos(x)+\cos(y)=2 \cos \left(\frac{x}{2}-\frac{y}{2}\right) \cos
   \left(\frac{x}{2}+\frac{y}{2}\right)$$
