How to evaluate from the number of solutions for $\sin 2x +\cos 3x - \sin 4x = 0$? The problem is as follows:

First find the number of solutions for the equation from below:
Assume $x \in [0,2\pi]$
$\sin 2x +\cos 3x - \sin 4x = 0$
Let $n$ be the number of solutions. Using this $n$ find the sum for:
$5\tan^2\left(\frac{n\pi}{18}\right)+1.5n^2$

I'm not sure how to solve this, what it came to my mind was this:
$\sin 2x +\cos 3x - \sin 4x = 0$
$\sin 4x - \sin 2x - \cos 3x = 0$
Using Prosthaphaeresis formulas then I'm getting to:
$\cos 3x \cdot (2\sin x - 1)=0$
From this it can be inferred that there will be six solutions for
$\cos 3x=0$
In the interval mentioned. Or at least I think so,
these will be:
$\frac{\pi}{6}$, $\frac{3\pi}{6}$, $\frac{5\pi}{6}$, $\frac{7\pi}{6}$, $\frac{9\pi}{6}$, $\frac{11\pi}{6}$.
While for the other guy it will be this:
$\sin x = \frac{1}{2}$
There will be two solutions in that interval, namely:
$\frac{\pi}{6}$, $\frac{5\pi}{6}$
Thus the number of solutions between the two would be eight. But when replacing in what it is being asked it doesn't seem to yield something reasonable.
Therefore can someone help me here?. Which part went wrong?.

I Think $n=6$.
Algebraically:
$$\sin 2x +\cos 3x -\sin 4x = \\ 2\sin x \cos x +\cos 2x \cos x-2\sin x \cos x \sin x -4 \sin x 
 \cos x \cos 2x= \\ 2 \sin x \cos x +(1-2 \sin^2 x)\cos x - 2 \sin^2 x \cos x -2 \sin x \cos x(2-4 \sin^2 x) = \\ 2 \sin x \cos x (1-2+4 \sin^2 x) + \cos x (1- 2 \sin^2 x - 2 \sin^2 x)= \\ 2\sin x \cos x(-1+4\sin^2 x) + \cos x (1-4 \sin^2 x)=  \\ 2\sin x \cos x(4\sin^2 x -1) - \cos x (4 \sin^2 x -1)= \\ (4 \sin^2 x -1)(\cos x)(2\sin x -1)=0 $$
so we have as solutions: $\frac{7\pi}{6},\frac{\pi}{6},\frac{5\pi}{6},\frac{11\pi}{6}$ for the first term, $\frac{\pi}{2},\frac{3\pi}{2}$ for the second, and $\frac{\pi}{6},\frac{5\pi}{6}$ for the third. Six solutions in total.
