Solve this trigonometric equation $\cos3x\cos x=\sin 3x$ Solve this equation $$ \cos 3x\cos x=\sin 3x$$
I tried converting product into sum but with no results. I think they forgot to add $\sin x$ after $\sin 3x$. 

Hint:
Notice, $$\cos 3x\cos x=\sin 3x$$ $$(4\cos^3x-3\cos x)\cos x=3\sin x-4\sin^3x$$
$$4\cos^4x-3\cos^2 x=3\sin x-4\sin^3x$$ $$3\sin x-4\sin^3x=4(1-\sin^2 x)^2-3(1-\sin^2x)$$   $$3\sin x-4\sin^3x=4+4\sin^4 x-8\sin^2x-3+3\sin^2x$$
$$4\sin^4 x+4\sin^3 x-5\sin^2x-3\sin x+1=0$$ Let, $\sin x=t$, we get $$4t^4+4t^3-5t^2-3t+1=0$$ 
I hope you can take it from here.
