I had to solve
$$\sin(5x) - \sin (3x) = \sqrt 2 \; \cos(4x)$$
After working with the equation, I got
$$2\sin(x)\cos(4x) = \sqrt 2 \; \cos(4x)$$
with difference of sines formula.
I saw that, I need to check if $\cos(4x)=0$ is a solution, so I got $x= \frac{\pi}{8}$ and it worked, so my first two solutions are
$$x= \frac{\pi}{8} + 2\pi n \space, n \in \Bbb Z \quad\text{and}\quad x= -\frac{\pi}{8} + 2\pi n \space, n \in \Bbb Z $$
After that, I divided by $\cos(4x)$, getting $\sin(x) = \frac{\sqrt 2}{2}$, so
$$x= \frac{\pi}{4} + 2\pi n \space, n \in \Bbb Z \quad\text{and}\quad x= \frac{3\pi}{4} + 2\pi n \space, n \in \Bbb Z $$
are another 2 solutions.
I checked in Wolfram Alpha to see if my work is correct, and I found that there are $10$ solutions to this equation. How can I get the other $6$ solutions I'm missing? (My 4 solutions are correct).