$10\sin(x)\cos(x) = 6\cos(x)$ 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
 A: Note that $\arcsin$ isn't an actual inverse of $\sin:\mathbb R\to[-1,1]$, but only one for
$$\sin: [-\frac\pi2,\frac\pi2]\to[-1,1]$$
Thus you must include solutions obtained by $\arcsin$ in all possible forms, that is
$$x_1 = \arcsin(\frac35),\qquad x_2 = \pi - \arcsin(\frac35)$$
And all solutions are of the form $2\pi k + x_j$ for $k\in\mathbb Z, j\in\{1,2\}$ plus $\pi k + \frac\pi2$ from the roots of $\cos$. In total:
$$10\sin x\cos x = 6\cos x\\
\Leftrightarrow x\in\{x_1 + 2\pi k, x_2 + 2\pi k, \frac\pi2 + \pi k | k\in\mathbb Z\}$$
A: $$10\sin x\cos x=6\cos x\iff 10\sin x\cos x-6\cos x=0\iff 2(\cos x)(5\sin x -3)=0\stackrel{:2}{\iff} (\cos x)(5\sin x -3)=0\iff \cos x=0 \; \; \lor \;\; 5\sin x-3=0$$
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 $\dfrac{\pi}{2}-\sin^{-1}\left (\dfrac35\right )$
A: Looks good to me, and it's good to see you made cases for $\cos(x) = 0$ and $\cos(x) \neq 0$. However, $\cos(x)=0$ whenever $x = \frac{(2n+1)\pi}{2}$ for all $n \in \Bbb{Z}$. This is different than what you have said, $x = \frac{\pi}{2}+2k\pi=\frac{(4k+1)\pi}{2}$. You will be missing half of the solutions that make $\cos(x)=0$ if you only consider  $x = \frac{\pi}{2}+2k\pi$
A: Clearly either $\sin x = \frac{3}{5}$ or $\cos x = 0$. Now you have to find out all the possible values for $x$.


*

*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 $x = \frac{\pi}{2} + k\pi$

