Find the general solution of $\cos(x)-\cos(2x)=\sin(3x)$ Problem:

Find the general solution of $$\cos(x)-\cos(2x)=\sin(3x)$$

I tried attempting this by using the formula$$\cos C-\cos D=-2\sin(\dfrac{C+D}{2})\sin(\dfrac{C-D}{2})$$
Thus, $$-2\sin\left(\dfrac{x}{2}\right)\sin\left(\dfrac{3x}{2}\right)=\sin 3x$$
$$\Rightarrow -2\sin\left(\dfrac{x}{2}\right)\sin\left(\dfrac{3x}{2}\right)-\sin 3x=0$$
Unfortunately, I couldn't get further with this problem. Any help  with this would be truly appreciated. Many thanks in advance!

*

*Expand the trig functions $$ \cos(x)+1-2 \cos^2(x) = 4 \sin(x) \cos^2(x)-\sin(x)$$

*Use the tangent half angle substitution $t=\tan(x/2)$, $\cos(x)=\frac{1-t^2}{1+t^2}$ and $\sin(x) = \frac{2 t}{1+t^2}$ $$ \frac{2 t^2 (3-t^2)}{(1+t^2)^2} = \frac{2 t (3 t^4-10 t^2+3)}{(1+t^2)^3}$$

*Collect terms $$\frac{2 t (t+1) (t^2-3) (t^2+2t-1)}{(1+t^2)^3} = 0$$

*Solve for $t$ $$\begin{align} t&=0\\t&=-1\\t&=\sqrt{3}\\t&=-\sqrt{3}\\t&=\sqrt{2}-1\\t&=-\sqrt{2}-1 \end{align}$$

*Solve for $x=2 \arctan(t)$ $$\begin{align}
x&=0\\
x&=-\frac{\pi}{2}\\
x&=\frac{2\pi}{3}\\
x&=-\frac{2\pi}{3}\\
x&=\frac{\pi}{4}\\
x&=-\frac{3\pi}{4}
\end{align}$$

