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