3. The original equation is equivalent to

$$
\begin{aligned}
\left(\frac{1-\cos 2 x}{2}\right)^{5}+\left(\frac{1+\cos 2 x}{2}\right)^{5}=\frac{29}{16} \cos ^{4} 2 x & \Longleftrightarrow \\
& \Longleftrightarrow \frac{2+20 \cos ^{2} 2 x+10 \cos ^{4} 2 x}{32}=\frac{29}{16} \cos ^{4} 2 x
\end{aligned}
$$

From which $24 \cos ^{4} 2 x-10 \cos ^{2} 2 x-1=0, \cos ^{2} 2 x=\frac{1}{2}, \cos 4 x=0$.
Answer: $x=\frac{\pi}{8}+\frac{\pi k}{4}$.

Answer to option: $5-2: x=\frac{\pi}{8}+\frac{\pi k}{4}$.