3.220

$$
\frac{1+\cos (2 \alpha-2 \pi)+\cos (4 \alpha+2 \pi)-\cos (6 \alpha-\pi)}{\cos (2 \pi-2 \alpha)+2 \cos ^{2}(2 \alpha+\pi)-1}=2 \cos 2 \alpha
$$
Solution.

$$
\begin{aligned}
& \frac{1+\cos (2 \alpha-2 \pi)+\cos (4 \alpha+2 \pi)-\cos (6 \alpha-\pi)}{\cos (2 \pi-2 \alpha)+2 \cos ^{2}(2 \alpha+\pi)-1}= \\
& =\frac{1+\cos (2 \pi-2 \alpha)+\cos (2 \pi+4 \alpha)-\cos (\pi-6 \alpha)}{\cos (2 \pi-2 \alpha)+2(\cos (\pi+2 \alpha))^{2}-1}= \\
& =\frac{1+\cos 2 \alpha+\cos 4 \alpha+\cos 6 \alpha}{\cos 2 \alpha+2 \cos ^{2} 2 \alpha-1}=\frac{1+\cos 2 \alpha+\cos 2(2 \alpha)+\cos 3(2 \alpha)}{2 \cos ^{2} 2 \alpha+\cos 2 \alpha-1}= \\
& =\left[\cos 2 x=2 \cos ^{2} x-1 ; \cos 3 x=4 \cos ^{3} x-3 \cos x\right]= \\
& =\frac{1+\cos 2 \alpha+2 \cos ^{2} 2 \alpha-1+4 \cos ^{3} 2 \alpha-3 \cos 2 \alpha}{2 \cos ^{2} 2 \alpha+\cos 2 \alpha-1}=
\end{aligned}
$$

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

The identity is proven.