3.290. $6 \sin ^{2} 2 \alpha-1-\cos 4 \alpha$.

3.290. $6 \sin ^{2} 2 \alpha-1-\cos 4 \alpha$.
Solution.

$$
\begin{aligned}
& 6 \sin ^{2} 2 \alpha-1-\cos 4 \alpha=6 \sin ^{2} 2 \alpha-1-\left(2 \cos ^{2} 2 \alpha-1\right)= \\
& =6 \sin ^{2} 2 \alpha-2 \cos ^{2} 2 \alpha=2\left(3 \sin ^{2} 2 \alpha-\cos ^{2} 2 \alpha\right)= \\
& =2(\sqrt{3} \sin 2 \alpha-\cos 2 \alpha)(\sqrt{3} \sin 2 \alpha+\cos 2 \alpha)= \\
& =2 \cdot 2\left(\frac{\sqrt{3}}{2} \sin 2 \alpha-\frac{1}{2} \cos 2 \alpha\right) \cdot 2\left(\frac{\sqrt{3}}{2} \sin 2 \alpha+\frac{1}{2} \cos 2 \alpha\right)=
\end{aligned}
$$

$=-8\left(\cos 2 \alpha \cos 60^{\circ}-\sin 2 \alpha \sin 60^{\circ}\right)\left(\cos 2 \alpha \cos 60^{\circ}+\sin 2 \alpha \sin 60^{\circ}\right)=$

$=[\cos x \cos y \pm \sin x \sin y=\cos (x \mp y)]=-8 \cos \left(2 \alpha+60^{\circ}\right) \cos \left(2 \alpha-60^{\circ}\right)$.

Answer: $-8 \cos \left(2 \alpha+60^{\circ}\right) \cos \left(2 \alpha-60^{\circ}\right)$.