3.292. $2 \cos ^{2} 2 \alpha+3 \cos 4 \alpha-3$.

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3.292. $2 \cos ^{2} 2 \alpha+3 \cos 4 \alpha-3$.
Solution.

$$
\begin{aligned}
& 2 \cos ^{2} 2 \alpha+3 \cos 4 \alpha-3=2 \cos ^{2} 2 \alpha+3\left(1-2 \sin ^{2} 2 \alpha\right)-3= \\
& =2 \cos ^{2} 2 \alpha-6 \sin ^{2} 2 \alpha=2\left(\cos ^{2} 2 \alpha-3 \sin ^{2} 2 \alpha\right)= \\
& =2(\cos 2 \alpha-\sqrt{3} \sin 2 \alpha)(\cos 2 \alpha+\sqrt{3} \sin 2 \alpha)= \\
& =2 \cdot 2\left(\frac{1}{2} \cos 2 \alpha-\frac{\sqrt{3}}{2} \sin 2 \alpha\right) \cdot 2\left(\frac{1}{2} \cos 2 \alpha+\frac{\sqrt{3}}{2} \sin 2 \alpha\right)= \\
& =8\left(\cos 2 \alpha \cos 60^{\circ}-\sin 2 \alpha \sin 60^{\circ}\right)\left(\cos 2 \alpha \cos 60^{\circ}+\sin 2 \alpha \sin 60^{\circ}\right)= \\
& =[\cos x \cos y \pm \sin x \sin y=\cos (x \mp y)]=8 \cos \left(2 \alpha+60^{\circ}\right) \cos \left(2 \alpha-60^{\circ}\right)
\end{aligned}
$$

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