Weird "hidden answer" in $2\tan(2x)=3\text{cot}(x)$ The question is

Find the solutions to the equation $$2\tan(2x)=3\cot(x) , \space 0<x<180$$

I started by applying the tan double angle formula and recipricoal identity for cot
$$2* \frac{2\tan(x)}{1-\tan^2(x)}=\frac{3}{\tan(x)}$$
$$\implies 7\tan^2(x)=3 \therefore x=\tan^{-1}\left(-\sqrt\frac{3}{7} \right)$$
$$x=-33.2,33.2$$
Then by using the quadrants

I was lead to the final solution that $x=33.2,146.8$ however the answer in the book has an additional solution of $x=90$, I understand the reasoning that $\tan(180)=0$ and $\cot(x)$ tends to zero as x tends to 90 however how was this solution found?
Is there a process for consistently finding these "hidden answers"?

Factorize the equation as follows
\begin{align}
2\tan(2x)-3\cot(x)
=& \frac{2\sin2x}{\cos 2x} - \frac{3\cos x}{\sin x}\\
=& \frac{2\sin2x\sin x-3 \cos x\cos2x }{ \sin x\cos 2x}\\
 =& \frac{\cos x(10\sin^2x-3 )}{ \sin x\cos 2x}\\ 
\end{align}
where the factor $\cos x =0$ captures the solution $x=\frac\pi2$ and $10\sin^2x -3=0$ yields $x= \sin^{-1}\sqrt{\frac3{10}}, \>\pi - \sin^{-1}\sqrt{\frac3{10}}$.
