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"?
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$\begingroup$$$\frac{4\tan(x)}{1-\tan^2(x)} = \frac{3}{\tan(x)}$$
$$\frac{4\tan(x)}{1-\tan^2(x)} - \frac{3}{\tan(x)} = 0$$
$$\frac{4\tan^2(x)-3[1-\tan^2(x)]}{\tan(x)[1-\tan^2(x)]} = 0$$
$$\frac{7\tan^2(x)-3}{\tan(x)[1-\tan^2(x)]} = 0$$
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).
$$\lim_{\tan(x)\to\infty} \frac{7\tan^2(x)-3}{\tan(x)[1-\tan^2(x)]} = \lim_{\tan(x)\to\infty} \frac{\tan^2(x)\left[7-\frac{3}{\tan^2(x)}\right]}{\tan^2(x)\left[\frac{1}{\tan(x)}-\tan(x)\right]} = \lim_{\tan(x)\to\infty} \frac{7-\frac{3}{\tan^2(x)}}{\frac{1}{\tan(x)}-\tan(x)} = 0$$
$\endgroup$ $\begingroup$The error is in your very first line.
As alluded to by @Tavish in the comments, since the double-angle identity for $\tan$ has domain $$\mathbb R\setminus \left\{(2k+1)\frac{\pi}2 \mid k\in\mathbb Z\right\},$$ a separate case for odd multiples of $90^{\circ}$ ought to be created when applying it. (In doing so while solving an equation where the flow of logic is $“\implies”$ rather than $“\iff”,$ we are merely asserting that this separate case contains potential solutions.)
So, for $x\in\left(0^{\circ},180^{\circ}\right),$
$$2\tan(2x)=3\cot(x) \tag1$$ $$\implies \frac{4\tan(x)}{1-\tan^2(x)}=\frac{3}{\tan(x)} \quad\text{or}\quad x=90^{\circ} \\ \implies \tan^2(x)=\frac37 \quad\text{or}\quad x=90^{\circ}$$ $$\implies x=33.2^{\circ}\:\text{or}\:146.8^{\circ} \quad\text{or}\quad x=90^{\circ}; \tag2 $$ plugging (2) into (1) reveals that the full solution set is$\left\{33.2^{\circ},90^{\circ},146.8^{\circ}\right\}.$
Alternatively, using “$\iff$”:
for $x\in\left(0^{\circ},180^{\circ}\right),$
$$2\tan(2x)=3\cot(x) \\ \iff \left[\frac{4\tan(x)}{1-\tan^2(x)}=\frac{3}{\tan(x)} \quad\text{or}\quad x=90^{\circ}\right] \quad\text{and}\quad \cos(2x),\sin(x)\neq0\\ \iff \left[\tan^2(x)=\frac37 \quad\text{or}\quad x=90^{\circ}\right] \quad\text{and}\quad \cos(2x),\sin(x)\neq0\\$$ $$\iff x=33.2^{\circ}\:\text{or}\: 90^{\circ}\:\text{or}\: 146.8^{\circ}.$$
In your second equation, the only way you can get from that equation to $7\tan^2{x} = 3$, is by assuming that $\tan{x}$ and $1 - \tan^2{x}$ is not equal to $0$ or infinity because you can't divide by $0$ or infinity. But the hidden case is that $1 - \tan^2{x} = 0$ or infinity or that $\tan{x} = 0$ or infinity.
We only see that one case works, and it is when $x = \boxed{90}$ degrees.
$\endgroup$ 4 $\begingroup$Perhaps this graph will help reveal the answers (abscissa in radians):
The moment you substitute $\cot x\mapsto \frac1{\tan x}$, you are implicitly assuming that $x\neq 90^\circ$, because that's required for that substitution to make sense. So that's a case you have to manually check in the original equation because it might be a solution that disappears (and in this case it indeed turned out to be a solution that disappears).
$\endgroup$ $\begingroup$Find the solutions to the equation
\begin{align} 2\tan(2x)=3\cot(x),\quad 0^\circ<x<180^\circ \tag{1}\label{1}\end{align}
As it was already noted, $\tan x$ is not defined on the whole range $(0^\circ,180^\circ)$, but$\cot x$ is, so if we use it instead in \eqref{1}:
\begin{align} \frac2{\cot 2x}&=3\cot(x) ,\\ \frac{4\cot x}{\cot^2 x-1} &= 3\cot x , \end{align}
we get the missing third solution, $\cot x=0$.
$\endgroup$ $\begingroup$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 $x= \sin^{-1}\sqrt{\frac3{10}}, \>\pi - \sin^{-1}\sqrt{\frac3{10}}$.
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