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?

So, we have ${\cos(2x + x) + \cos(x) = 0}$. Using the compound trig formulae:
$${\Rightarrow \cos(2x)\cos(x) - \sin(2x)\sin(x) + \cos(x) = 0}$$
Using the double angle identity for ${\sin(2x)}$, and ${\cos(2x)}$ gives
$${(2\cos^2(x) - 1)\cos(x) - 2\cos(x)\sin^2(x) + \cos(x)= 0}$$
$${\Rightarrow 2\cos^3(x) - 2\cos(x)\left(1 - \cos^2(x)\right)=0}$$
$${\Rightarrow 4\cos^3(x) - 2\cos(x) = 0}$$
$${\Rightarrow \cos(x)\left(4\cos^2(x) - 2\right)=0}$$
So either ${\cos(x)=0}$ or ${4\cos^2(x) - 2 = 0}$. Can you take it from here?
