![5\sin2x=3\cos x\iff10\sin x\cos x=3\cos x](/tpl/images/0469/7339/32a49.png)
by the double angle identity for sine. Move everything to one side and factor out the cosine term.
![10\sin x\cos x=3\cos x\iff10\sin x\cos x-3\cos x=\cos x(10\sin x-3)=0](/tpl/images/0469/7339/b1a0a.png)
Now the zero product property tells us that there are two cases where this is true,
![\begin{cases}\cos x=0\\10\sin x-3=0\end{cases}](/tpl/images/0469/7339/1ed53.png)
In the first equation, cosine becomes zero whenever its argument is an odd integer multiple of
![\dfrac\pi2](/tpl/images/0469/7339/65196.png)
, so
![x=\dfrac{(2n+1)\pi}2](/tpl/images/0469/7339/737ca.png)
where
![n[/tex ]is any integer.\\Meanwhile,\\[tex]10\sin x-3=0\implies\sin x=\dfrac3{10}](/tpl/images/0469/7339/f32ec.png)
which occurs twice in the interval
![[0,2\pi)](/tpl/images/0469/7339/8667d.png)
for
![x=\arcsin\dfrac3{10}](/tpl/images/0469/7339/0aa69.png)
and
![x=\pi-\arcsin\dfrac3{10}](/tpl/images/0469/7339/407c9.png)
. More generally, if you think of
![x](/tpl/images/0469/7339/a0e3f.png)
as a point on the unit circle, this occurs whenever
![x](/tpl/images/0469/7339/a0e3f.png)
also completes a full revolution about the origin. This means for any integer
![n](/tpl/images/0469/7339/a4ae7.png)
, the general solution in this case would be
![x=\arcsin\dfrac3{10}+2n\pi](/tpl/images/0469/7339/6d0c7.png)
and
![x=\pi-\arcsin\dfrac3{10}+2n\pi](/tpl/images/0469/7339/4e8ab.png)
.