Trigonometry equation that includes two sin and a cos

Question

How would I arrange this equation $(3\sin2x=5\sin 2x\cos2x)$ to get either $\cos \sin$ or $\tan$ by itself?

I’ve tried to do the $\sin/\cos$ to get $\tan$, but I’m unsure how to get the second $\sin$ in terms of $\tan$ or if I am going in the wrong direction Thank you!!! :)

Answer 1

$$3\sin 2x=5\sin2x\cos 2x\implies \sin 2x=0 \land \cos 2x=\dfrac{3}{5}$$

$$x=\displaystyle\bigcup_{k=0}^{n}\left(\dfrac{\pi k}{2}\right)\cup {\dfrac{1}{2}\arccos\dfrac{3}{5}}$$

Answer 2

Write your equation in the form. $$\sin(2x)(3-5\cos(2x))=0$$