Firstly, notice that \(\displaystyle{\tan{{\left({2}\theta\right)}}}\) appears in both terms on the left so let's factor that out.

\(\displaystyle{\tan{{\left({2}θ\right)}}}{\left[{1}−{{\tan}^{{2}}{\left(θ\right)}}\right]}={2}\)

Now, we need the arguments of our tantan functions to be the same if we have any hopes of simplifying the left-hand side. So we'll use the double angle formula to convert that \(\displaystyle{\tan{{\left({2}θ\right)}}}\) into an expression in terms of \(\tan(θ)\). So remember that the double angle formula for tantan is

\(\displaystyle{\tan{{\left({2}θ\right)}}}=\frac{{{2}{\tan{{\left(θ\right)}}}}}{{{1}-{\left({{\tan}^{{2}}{\left(θ\right)}}\right)}}}\)

Using that, we get \(\displaystyle\to{\tan{{\left(θ\right)}}}={1}\)

Here we look at the unit circle. Note that \(\displaystyle{\tan{{\left(θ\right)}}}={1}\) means the same thing as \(\displaystyle{\sin{{\left(θ\right)}}}={\cos{{\left(θ\right)}}}\) which means the same thing as the x component = the y component on the unit circle. That only occurs at two angles between 0 and \(\displaystyle{2}\pi\) radians:\(\displaystyle\frac{\pi}{{4}}\ \text{ radians }\ \ \text{ and }\ {5}\frac{\pi}{{4}}\) radians.

Remember though that the angle doesn't necessarily have to be between 0 and \(2\pi\) radians so we need to add \(2\pi k\) to both of those angles. Hence the full solution is \(\frac{\pi}{4}+\frac{2}{\pi k}\) or \(\frac{5\pi}{4+2\pi k}\) for every integer k, or more succinctly

\(\frac{\pi}{4}+\pi k\) for every integer k