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"?

$$\frac{4\tan(x)}{1-\tan^2(x)} = \frac{3}{\tan(x)}$$
$$\frac{4\tan(x)}{1-\tan^2(x)} - \frac{3}{\tan(x)} = 0$$
$$\frac{4\tan^2(x)-3[1-\tan^2(x)]}{\tan(x)[1-\tan^2(x)]} = 0$$
$$\frac{7\tan^2(x)-3}{\tan(x)[1-\tan^2(x)]} = 0$$
You focused in the fact that the equation is satisfied when the numerator is zero, i.e., $7\tan^2(x)-3=0$, but the equation is also satisfied when $\tan(x)\to\infty$ (when the denominator itself tends to infinity).
$$\lim_{\tan(x)\to\infty} \frac{7\tan^2(x)-3}{\tan(x)[1-\tan^2(x)]} = \lim_{\tan(x)\to\infty} \frac{\tan^2(x)\left[7-\frac{3}{\tan^2(x)}\right]}{\tan^2(x)\left[\frac{1}{\tan(x)}-\tan(x)\right]} = \lim_{\tan(x)\to\infty} \frac{7-\frac{3}{\tan^2(x)}}{\frac{1}{\tan(x)}-\tan(x)} = 0$$
