Weierstrass substitution on an algebraic expression I start with the trigonometric equation $3\tan{2x} + 16\cot^2{x} = 0$, although there is a particular solution pathway that I can follow by taking the double-angle identity to reduce the LHS to a cubic expression in terms of $\tan{x}$, I'm curious on a particular line of reasoning that follows by substituting $t = \tan(\frac{x}{2})$, and substituting the typical expressions after applying $$\tan{2x} = \frac{2\tan{x}}{1 - \tan^2{x}},~~\cot^2{x} = \frac{1}{\tan^2{x}},$$ where $\tan{x} = \frac{2t}{1-t^2}$.
My two questions posed are:
(a) Is the Weierstrass substitution here valid in this context of manipulating trigonometric equations algebraically?
(b) If so, is this particular case worthwhile trying to solve using such a substitution?
Post-note: A better question would be whether such substitutions are even well-formed in algebra, there is the constraint that $0 \leq x \leq \pi$, so I would've presumed there'd be no conflicts between domains/ranges if that is even a consideration.
 A: The Weierstrass substitution is always valid in whatever you're doing, be it integrating tricky trigonometric integrals or solving trig equations (eg the one you're solving). This is because the Weierstrass substitution $t=\tan\frac{x}{2}$ relies on identities, and identities are by definition always true.
However, as noted in the comments, the Weierstrass substitution can introduce extra solutions into the equation we are solving given the constraints you are dealing with, so if you use it always doublecheck if the solutions you obtained make sense.
In this particular case, without using the Weierstrass substitution we have the cubic equation
$$\frac{6\tan x}{1-\tan^2 x}+\frac{16}{\tan^2x}=0\iff3\tan^3x-8\tan^2x+8=0$$
By using the factor theorem and trying small values for $\tan x$, I quite quickly saw that $\tan x=2$ is a solution and therefore $\tan x-2$ is a factor of our expression. Hence, our expression is equivalent to
$$(\tan x-2)(3\tan^2x-2\tan x-4)=0$$
And the second factor is just a quadratic in $\tan x$ which we can solve easily.
So, all in all, in this case my personal preference would be not to use the Weierstrass substitution; it would've complicated an expression which turned out to be rather nice.
On the other hand, eg if we had the equation $3\sin x+2\cos x-9=0$ then the Weierstrass substitution is very useful if we don't want to resort to the $R\cos(x+\alpha)$ and $R\sin(x+\alpha)$ formulae (indeed I very rarely want to resort to them). In general, if we have the equation $a\sin x+b\cos x+c=0$ then I always recommend using the Weierstrass substitution, unless it's something really simple.

I hope that was helpful. If you have any questions please don't hesitate to ask :)
A: The substitution is perfectly valid, but it can miss some solutions. If we consider
$$
\tan2x=\frac{2\tan x}{1-\tan^2x}\tag{*}
$$
we see that the left-hand side is defined for $x=\pi/2$, whereas the right-hand side isn't. Same problem for $x=3\pi/2$. The right-hand side is also undefined for $x=\pi/4+k\pi$ and $x=3\pi/4+k\pi$, but so isn't the left-hand side as well.
Thus, when you apply the identity to an equation, you also have to consider the possibly missed solutions. In your case, $x=\pi/2+k\pi$ is a solution!
But you may also multiply numerator and denominator the right-hand side in (*) by $\cot^2x$ so obtaining
$$
\tan2x=\frac{2\cot x}{\cot^2x-1}\tag{**}
$$
which isn't free from the same issue, because $x=k\pi$ makes the right-hand side invalid. However, in the particular case, this is not of a concern, because the equation also contains $\cot x$.
Upon applying the identity (**), you get
$$
\frac{6\cot x}{\cot^2x-1}+16\cot^2x=0
$$
and you can factor out $2\cot x$, after observing that it gives the solutions $x=\pi/2+k\pi$.
Thus the equation becomes
$$
3+8\cot^3x-8\cot x=0
$$
that factors as
$$
(2\cot x-1)(4\cot^2x+2\cot x-3)=0
$$
