3.307. $2 \sin ^{2} 2 \alpha+\sqrt{3} \sin 4 \alpha-\frac{4 \tan 2 \alpha\left(1-\tan^{2} 2 \alpha\right)}{\sin 8 \alpha\left(1+\tan^{2} 2 \alpha\right)^{2}}$.
Solution.

$$
\begin{aligned}
& 2 \sin ^{2} 2 \alpha+\sqrt{3} \sin 4 \alpha-\frac{4 \operatorname{tg} 2 \alpha\left(1-\operatorname{tg}^{2} 2 \alpha\right)}{\sin 8 \alpha\left(1+\operatorname{tg}^{2} 2 \alpha\right)^{2}}= \\
& =2 \sin ^{2} 2 \alpha+\sqrt{3} \sin 4 \alpha-\frac{2 \operatorname{tg} 2 \alpha}{1+\operatorname{tg}^{2} \alpha} \cdot \frac{2\left(1-\frac{\sin ^{2} 2 \alpha}{\cos ^{2} 2 \alpha}\right)}{\left(1+\frac{\sin ^{2} 2 \alpha}{\cos ^{2} 2 \alpha}\right) \sin 2(4 \alpha)}= \\
& =\left[\sin ^{2} \frac{x}{2}=\frac{1-\cos x}{2}, \frac{2 \operatorname{tg} \frac{x}{2}}{1+\operatorname{tg}^{2} \frac{x}{2}}=\sin x\right]= \\
& =1-\cos 4 \alpha+\sqrt{3} \sin 4 \alpha-\frac{\cos ^{2} 2 \alpha+\sin ^{2} 2 \alpha}{\sin ^{2} 2 \alpha} \cdot 2 \sin 4 \alpha \cos 4 \alpha \\
& =1-\cos 4 \alpha+\sqrt{3} \sin 4 \alpha-\frac{\cos ^{2} 4 \alpha}{\cos ^{2} \alpha}=\sqrt{3} \sin 4 \alpha-\cos 4 \alpha= \\
& =2\left(\frac{\sqrt{3}}{2} \sin 4 \alpha-\frac{1}{2} \cos 4 \alpha\right)=2\left(\sin ^{2} 4 \alpha \cos \frac{\pi}{6}-\cos 4 \alpha \sin \frac{\pi}{6}\right)= \\
& =[\sin x \cos y-\cos x \sin y=\sin (x-y)]=2 \sin \left(4 \alpha-\frac{\pi}{6}\right) \\
& \text { Result }: 2 \sin \left(4 \alpha-\frac{\pi}{6}\right) \cdot
\end{aligned}
$$