How do you solve for $\alpha$ in this trigonometric equation? I have $$\tan (2\alpha) = \frac {4n^2}{4n^4-1}$$
And I want to solve for $\alpha$. So far I have tried applying the inverse tangent to both sides and dividing by two, but the book says that the answer should be $\frac {1}{2n^2}$ .

So then I use the double angle formulas for sine and cosine and after some rearraging I get to this:$$2n^2-\frac {1}{2n^2}=\cot(\alpha) -\tan(\alpha)$$ Which seems to imply what the book says but I am unable to get rid of the cotangent to get a rigorous result. How should I attack the problem?
 A: $$
\underbrace{\frac {2\tan\alpha}{1-\tan^2\alpha} = \tan(2\alpha)}_\text{double-angle formula} = \underbrace{\frac{4n^2}{4n^4-1} = \frac{2\left( \dfrac 1 {2n^2} \right)}{1 - \left(\dfrac 1{2n^2}\right)^2}}_{\begin{smallmatrix} \text{This gets us a “1'' where we} \\  \text{need it in the denominator.} \end{smallmatrix}}.
$$
We have $\dfrac{2\tan\alpha}{1-\tan^2\alpha} = \dfrac{2\left( \dfrac 1 {2n^2} \right)}{1 - \left(\dfrac 1{2n^2}\right)^2}\quad$ if $\quad\tan\alpha = \dfrac 1{2n^2}$.
PS in response to comments: Two issues arise.
First, if $\dfrac{2x}{1-x^2} = \dfrac{2y}{1-y^2}$, does it follow that $x=y$? If not, we could have $\tan\alpha=x$ and $\dfrac 1 {2n^2} = y$; thus where I said $\text{“}$If $x=y\text{''}$, one could not then add $\text{“}$only if$\text{''}$.  Then one would have to check for extraneous roots.  One way to deal with this is to actually check for extraneous roots.
In fact the equation $\dfrac{2x}{1-x^2} = \dfrac{2y}{1-y^2}$ is equivalent to
$$ 2x(1-y^2) = 2y(1-x^2) \text{ and } x,y\notin\{-1,1\}, $$
unless you want to allow $\infty$ as a value (which might not be a bad idea) in which case you can drop the part including and after the word $\text{“and''}$.  This is a quadratic equation and if you solve it for $x$ in terms of $y$ (or vice-versa) you do get two distinct solutions.  Howeover, both of the resulting distinct values of $\alpha$ lead to the same value of $2\alpha$, and therefore to a solution of the original equation.
Second: Do we only want solutions in the interval $(-\pi/2,\pi/2)$?  If so, then you don't have to list infinitely many solutions; otherwise you do.
A: You have $$\tan2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha}=\frac{4n^2}{4n^4-1}$$
Rearranging this gives $$\tan^2\alpha+\tan\alpha\left(2n^2-\frac{1}{2n^2}\right)-1=0$$
$$\Rightarrow\left(\tan\alpha+2n^2\right)\left(\tan\alpha-\frac{1}{2n^2}\right)=0$$
Hence $$\tan\alpha=-2n^2 \text{or} \frac{1}{2n^2}$$
