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 $\sin 3x$. 

We have$$\begin{align}\\&\cos 3x\cos x=\sin 3x\\&\iff (4\cos^3x-3\cos x)\cos x=3\sin x-4\sin^3x\\&\iff 4\cos^4x-3\cos^2x=3\sin x-4\sin^3x\\&\iff 4(1-\sin^2x)^2-3(1-\sin^2x)=3\sin x-4\sin^3x\end{align}$$
Here, let $t=\sin x$.
$$\begin{align}\\& 4(1-t^2)^2-3(1-t^2)=3t-4t^3\\&\iff 4(1-2t^2+t^4)-3+3t^2-3t+4t^3=0\\&\iff 4-8t^2+4t^4-3+3t^2-3t+4t^3=0\\&\iff 4t^4+4t^3-5t^2-3t+1=0\\&\iff 16t^4+16t^3-20t^2-12t+4=0\\&\iff (4t^2+2t-3)^2-5=0\\&\iff (4t^2+2t-3-\sqrt 5)(4t^2+2t-3+\sqrt 5)=0\\&\iff \sin x=t=\frac{-1\pm\sqrt{13+4\sqrt 5}}{4},\frac{-1\pm\sqrt{13-4\sqrt 5}}{4}\end{align}$$
Here, note that 
$$\frac{-1-\sqrt{13+4\sqrt 5}}{4}\lt -1.$$
