Really want this one solved

Here's a helping hand:

See if you can take it from here.

because of DeMoivre's Theorem:

$\begin{align*} \operatorname{cis} n \theta &= (\operatorname{cis} \theta)^n \\ &= (\cos \theta + i \sin \theta)^n \\ &= \cos^n \theta + \binom{n}{1} i \cos^{n - 1} \theta \sin \theta - \binom{n}{2} \cos^{n - 2} \theta \sin^2 \theta - \binom{n}{3} i \cos^{n - 3} \theta \sin^3 \theta + \dotsb. \end{align*}$

$\begin{align*} \cos n \theta &= \cos^n \theta - \binom{n}{2} \cos^{n - 2} \theta \sin^2 \theta + \binom{n}{4} \cos^{n - 4} \theta \sin^4 \theta - \dotsb, \\ \sin n \theta &= \binom{n}{1} \cos^{n - 1} \theta \sin \theta - \binom{n}{3} \cos^{n - 3} \theta \sin^3 \theta + \binom{n}{5} \cos^{n - 5} \theta \sin^5 \theta - \dotsb. \end{align*}$

$\begin{align*} \sin 5 \theta &= \binom{5}{1} \cos^4 \theta \sin \theta - \binom{5}{3} \cos^2 \theta \sin^3 \theta + \binom{5}{5} \sin^5 \theta \\ &= 5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta. \end{align*}$

$\sin 5 \theta = \sin^5 \theta$ 

becomes 

$5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta = \sin^5 \theta$

$5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta = 0$

$\theta$ is an acute angle, so $\cos^2 \theta - 2 \sin^2 \theta = 0$

$\cos^2 \theta = 2 \sin^2 \theta$

so, I know that $\tan^2 \theta = \frac{1}{2}$

$\tan \theta = \frac{1}{\sqrt{2}}$

$\tan 2 \theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} = \frac{\sqrt{2}}{1 - \frac{1}{2}} = {2 \sqrt{2}}$