Finding general solutions of trigonometric equations I have to find general solutions for $$\cos(x) + \cos(2x) = 0.$$ From sum-to-product formula I got $$\cos(3x/2) \cos(x/2) = 0,$$ giving me $$x = \pi/3$$ and $$x = \pi.$$  According to my text book, the answer should be in the form $$2n\pi \pm \alpha.$$ So I gave the answer $$2n\pi \pm \frac\pi3.$$ But the answer in the back of the book has left the $$\pm$$ sign and it says $$(2n + 1)\frac\pi3.$$  I need to know if $$\pm$$ sign in my answer is unnecessary or not.
 A: The $\pm$ sign in your answer is not needed. Think about the solutions that you are seeking as a set. You are claiming that the set of all the solutions is the set 
$$
\bigl\{(2n-1)\tfrac{\pi}{3}, (2n+1)\tfrac{\pi}{3} \ | \ n \in \mathbb Z \bigr\}. 
$$
Here $\mathbb Z$ denotes the set of all integers. The last set is the union of the following two sets 
$$
\bigl\{(2n-1)\tfrac{\pi}{3}\ | \ n \in \mathbb Z \bigr\} \quad \text{and} \quad \bigl\{(2n+1)\tfrac{\pi}{3} \ | \ n \in \mathbb Z \bigr\}. 
$$
However, it is not difficult to show that 
$$
\bigl\{(2n-1)\tfrac{\pi}{3}\ | \ n \in \mathbb Z \bigr\} = \bigl\{(2n+1)\tfrac{\pi}{3} \ | \ n \in \mathbb Z \bigr\}. 
$$
For example $-\pi$ belongs to the set on the left-hand side since $-\pi = (2(-1)-1)\tfrac{\pi}{3}$ and $-\pi$ belongs to the set on the right-hand side since $-\pi = (2(-2)+1)\tfrac{\pi}{3}$. In general, $(2n-1)\tfrac{\pi}{3} = (2(n-1)+1)\tfrac{\pi}{3}$; so each element in the set on the left-hand side is in the set on the right-hand side. Similarly, $(2n+1)\tfrac{\pi}{3} = (2(n+1)-1)\tfrac{\pi}{3}$; so each element in the set on the right-hand side is in the set on the left-hand side. 
So, the correct answer is also 
$$
(2n-1)\tfrac{\pi}{3}, \quad n \in \mathbb Z.
$$
A: The idea here is that you want to see if the range from $(2n+1)$ if different from that of $(2n\pm1)$.
Because $y = cos(x) + cos(2x)$ has constant periodicity, you can assume $n \in \mathbb{Z}$.
The range of $(2n + 1)$ will be $\mathbb{Z}$.
The range of $(2n \pm 1)$ also will be $\mathbb{Z}$.
Therefore, your answer yields the same result as the one from your textbook.
