<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] Clearly either $\sin x = \frac{3}{5}$ or $\cos x = 0$. Now you have to find out all the possible values for $x$.


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*If $\sin x = \frac{3}{5}$, then $x=\arcsin \frac{3}{5} + 2k\pi$ ... or $x = \pi - \arcsin \frac{3}{5} + 2k\pi$.

*If $\cos x = 0$, then $x = \frac{\pi}{2} + 2k\pi$ ... or $x = -\frac{\pi}{2} + 2k\pi$. This is the same as $x = \frac{\pi}{2} + k\pi$

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