3.286. $\frac{1}{\sqrt{3}} \sin 4 \alpha + 1 - 2 \cos ^{2} 2 \alpha$.
Solution.

$$
\frac{1}{\sqrt{3}} \sin 4 \alpha + 1 - 2 \cos^2 2 \alpha = \frac{1}{\sqrt{3}} \sin 4 \alpha - (2 \cos^2 2 \alpha - 1) =
$$

$= \frac{1}{\sqrt{3}} \sin 4 \alpha - \cos 4 \alpha = \frac{2}{\sqrt{3}} \left( \frac{1}{2} \sin 4 \alpha - \frac{\sqrt{3}}{2} \cos 4 \alpha \right) =$

$= \frac{2}{\sqrt{3}} \left( \cos 60^\circ \sin 4 \alpha - \sin 60^\circ \cos 4 \alpha \right) =$

$= \frac{2}{\sqrt{3}} \left( \sin 4 \alpha \cos 60^\circ - \cos 4 \alpha \sin 60^\circ \right) =$

$= [\sin x \cos y - \cos x \sin y = \sin (x - y)] = \frac{2}{\sqrt{3}} \sin \left( 4 \alpha - 60^\circ \right)$.

Answer: $\frac{2}{\sqrt{3}} \sin \left( 4 \alpha - 60^\circ \right)$.