<s>[INST] Solving trigonometric equations: $\cos(3x)+ \cos(x)=0$ Recently I worked on a problem where I had to solve $$\cos(3x)+\cos(x)=0$$
When I tried calculating it  by evaluating $x$, $x=2nπ+\dfrac{\pi}{4}$, $x=2n\pi-\dfrac{\pi}{4}$, or $x=2n\pi+\dfrac{\pi}{2}$ was the solution I reached.
Unfortunately, it was apparently wrong. Is there another way to solve this problem?
 [/INST] Symmetry is always a powerful and nice instrument
$$
\eqalign{
  & 0 = \cos (3x) + \cos x =   \cr 
  &  = \cos (2x + x) + \cos (2x - x) =   \cr 
  &  = \cos (2x)\cos x - \sin (2x)\sin x + \cos (2x)\cos x + \sin (2x)\sin x =   \cr 
  &  = 2\cos (2x)\cos x \cr} 
$$
and then
$$
\eqalign{
  & \cos (2x) = 0\quad  \vee \quad \cos x = 0  \cr 
  & 2x = {\pi  \over 2} + k\pi \quad  \vee \quad x = {\pi  \over 2} + j\pi   \cr 
  & x = {\pi  \over 4} + k{\pi  \over 2}\quad  \vee \quad x = {\pi  \over 2} + j\pi   \cr 
  & x = {\pi  \over 4} + 2k{\pi  \over 4}\quad  \vee \quad x = 2{\pi  \over 4} + 4j{\pi  \over 4}  \cr 
  & x = \left( {2k + 1} \right){\pi  \over 4}\quad  \vee \quad \left( {2j + 1} \right){\pi  \over 2}\quad \left| {\,k,j \in Z} \right. \cr} 
$$
</s>