# Trigonometry equation that includes two sin and a cos

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!!! :)

## migrated from mathematica.stackexchange.comApr 1 at 10:38

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• Simplify the equation ! – Yves Daoust Apr 1 at 10:43

Write your equation in the form. $$\sin(2x)(3-5\cos(2x))=0$$
$$3\sin 2x=5\sin2x\cos 2x\implies \sin 2x=0 \land \cos 2x=\dfrac{3}{5}$$
$$x=\displaystyle\bigcup_{k=0}^{n}\left(\dfrac{\pi k}{2}\right)\cup {\dfrac{1}{2}\arccos\dfrac{3}{5}}$$