[tex]5\sin2x=3\cos x\iff10\sin x\cos x=3\cos x[/tex]
by the double angle identity for sine. Move everything to one side and factor out the cosine term.
[tex]10\sin x\cos x=3\cos x\iff10\sin x\cos x-3\cos x=\cos x(10\sin x-3)=0[/tex]
Now the zero product property tells us that there are two cases where this is true,
[tex]\begin{cases}\cos x=0\\10\sin x-3=0\end{cases}[/tex]
In the first equation, cosine becomes zero whenever its argument is an odd integer multiple of [tex]\dfrac\pi2[/tex], so [tex]x=\dfrac{(2n+1)\pi}2[/tex] where [tex]n[/tex ]is any integer.
Meanwhile,
[tex]10\sin x-3=0\implies\sin x=\dfrac3{10}[/tex]
which occurs twice in the interval [tex][0,2\pi)[/tex] for [tex]x=\arcsin\dfrac3{10}[/tex] and [tex]x=\pi-\arcsin\dfrac3{10}[/tex]. More generally, if you think of [tex]x[/tex] as a point on the unit circle, this occurs whenever [tex]x[/tex] also completes a full revolution about the origin. This means for any integer [tex]n[/tex], the general solution in this case would be [tex]x=\arcsin\dfrac3{10}+2n\pi[/tex] and [tex]x=\pi-\arcsin\dfrac3{10}+2n\pi[/tex].
