Trigonometric equation: $\cos3x+\cos x-\cos2x=0$ $\cos3x+\cos x-\cos2x=0$ find general solution
My answer:
$\cos 3x+\cos x-\cos2x=0$
$\cos2x(2\cos x-1)=0$
$\cos2x=0$ or $2\cos x-1=0$
When $\cos2x=0$
$x=(n+1)\pi/2$
When $\cos x=1/2$
Solving...$x=\pi/3$
Then $x=2n\pi\pm \pi/3$
Did I go wrong?
Is it correct?
 A: $$\cos(x)-\cos(2x)+\cos(3x)=0\Longleftrightarrow$$
$$1-2\cos(x)-2\cos^2(x)+4\cos^3(x)=0\Longleftrightarrow$$
$$\left(2\cos(x)-1\right)\cdot\left(2\cos^2(x)-1\right)=0\Longleftrightarrow$$
$$2\cos(x)-1=0\Longleftrightarrow \vee 2\cos^2(x)-1=0\Longleftrightarrow$$
$$2\cos(x)=1\Longleftrightarrow \vee 2\cos^2(x)=1\Longleftrightarrow$$
$$\cos(x)=\frac{1}{2}\Longleftrightarrow \vee \cos^2(x)=\frac{1}{2}\Longleftrightarrow$$
$$\cos(x)=\frac{1}{2}\Longleftrightarrow \vee \cos(x)=\pm\frac{1}{\sqrt{2}}\Longleftrightarrow$$
$$\cos(x)=\frac{1}{2}\Longleftrightarrow \vee \cos(x)=\frac{1}{\sqrt{2}} \vee \cos(x)=-\frac{1}{\sqrt{2}} \Longleftrightarrow$$
$$x=\frac{\pi}{3}+2\pi n_1 \vee x=\frac{5\pi}{3}+2\pi n_2 \vee x=\frac{\pi}{4}+2\pi n_3 \vee x=\frac{7\pi}{4}+2\pi n_4$$
$$\vee x=\frac{3\pi}{4}+2\pi n_5 \vee x=\frac{5\pi}{4}+2\pi n_6$$
With $n_1,n_2,n_3,n_4,n_5,n_6 \in \mathbb{Z}$
A: You're going to have to do a lot more work.
First, find out what the identities are for $\cos 3 \theta$ and $\cos 2 \theta$ in terms of $\cos \theta$.
It looks like you know how to factorise - that will be important and you are doing the right sort of thing with your factors.
But you need to know how to express a "general solution" rather than just a "principal solution" to an equation involving trigonometrical ratios.
For example, to solve $\sin x=0.5$ I can just type "$\sin^{-1}(0.5)$ into my calculator and I will get the result $30^{\circ}$ (the principal value) but there are many other angles that also have $\sin x=0.5$: $150^{\circ}, 390^{\circ}$ etc. You need to know how to express those more generally.
A: If you know the formula the formula 
$$\cos\alpha+\cos\beta=2\cos\frac{\alpha+\beta}2\cos\frac{\alpha-\beta}2,$$ you can use it to get
$$\cos 3x+\cos x=2\cos2x \cos x.$$
This should simplify the original problem significantly.
From what you wrote in the question, it seems that you probably wanted to use this.
