<s>[INST] $10\sin(x)\cos(x) = 6\cos(x)$ In order to solve 
$$10\sin x\cos x = 6\cos x$$
I can suppose: $\cos x\ne0$ and then:
$$10\sin x = 6\implies \sin x = \frac{6}{10}\implies x = \arcsin \frac{3}{5}$$
And then, for the case $\cos x = 0$, we have:
$$x = \frac{\pi}{2} + 2k\pi$$
Is my solution rigth? Because I'm getting another strange things at wolfram alpha
 [/INST] $$10\sin x\cos x=6\cos x\iff 10\sin x\cos x-6\cos x=0\iff 2(\cos x)(5\sin x -3)=0\stackrel{:2}{\iff} (\cos x)(5\sin x -3)=0\iff \cos x=0 \; \; \lor \;\; 5\sin x-3=0$$
Is it any more clear now? Also, that $\cos x=0$ in your solution has to be $x=\dfrac{\pi}{2}+k\pi\, ,k\in \Bbb Z$ and your $\sin x=0$ has to include solutions in the form $\dfrac{\pi}{2}-\sin^{-1}\left (\dfrac35\right )$
</s>