Resolve $\cos(3x)= \cos(2x)$ I have to solve $$\cos(3x)= \cos(2x)$$
I found how to express both of them for $x$ only and now I have
$$4\cos^3(x)-3\cos(x)=2\cos^2(x)-1$$
What do I do now?
 A: We have the formula
$$ \cos{(a+b)}-\cos{(a-b)} = -2\sin{a}\sin{b}, $$
so if we put $a=5x/2,b=x/2$, we have
$$ 0 = \cos{3x}-\cos{2x} = -2\sin{\left(\frac{5}{2}x\right)}\sin{\left(\frac{x}{2}\right)},$$
so you just have to determine the $x$ for which either one of the sines is equal to $0$, which I'm sure you can do.
A: Hint: 
Use this result
$$\cos(a)=\cos(b)\iff a\equiv \pm b\pmod{2\pi}$$
A: You reached to the cubic equation $$4t^3-2t^2-3t+1=0$$ where $$t=\cos(x)\in [-1,1]$$
One root is $t=1$. Divide the polynomial by $(t-1)$ and find the other two roots. Finally plug in $t=\cos(x)$ to find $x$.
A: Notice, we have $$\cos(3x)=\cos(2x)$$ 
$$3x=2n\pi\pm 2x$$
Where, $n$ is any integer 
Now, we have the following solutions 
$$3x=2n\pi+2x\implies \color{red}{x=2n\pi}$$
or 
$$3x=2n\pi-2x\implies \color{red}{x=\frac{2n\pi}{5}}$$
A: hint: $$\cos \alpha=\cos \beta$$$$\iff \alpha=2k\pi\pm \beta$$
where; $k=0, \pm1, \pm2, \pm 3, \ldots$
A: $$\cos(3x)=\cos(2x)$$
$$\cos(3x)-\cos(2x)=0$$
$$2\sin(3x/2)\sin(-x/2)=0$$
$$-2\sin(3x/2)\sin(x/2)=0$$
$$\sin(3x/2)=0\implies 3x/2=k\pi\iff x=2k\pi/3$$
or $$\sin(x/2)=0\implies x/2=k\pi\iff x=2k\pi$$
