4. Determine the minimum value of the quantity $|x+y|$ given that the numbers $x$ and $y$ satisfy the relation $5 \cos (x+4 y)-3 \cos (x-4 y)-4 \sin (x-4 y)=10$.
Solution. Let $\varphi=\arccos \frac{3}{5}$ and we get $\cos (x+4 y)-\cos (x-4 y-\varphi)=2 \Leftrightarrow$ $\left\{\begin{array}{c}\cos (x+4 y)=1, \\ \cos (x-4 y-\varphi)=-1 .\end{array}\right.$ Therefore, $\left\{\begin{array}{c}x=\frac{\varphi+\pi}{2}+\pi(n+k), \\ y=-\frac{\varphi+\pi}{8}+\frac{\pi(n-k)}{4}\end{array} \Rightarrow\right.$ $x+y=\frac{3 \varphi+\pi}{8}+\frac{\pi(5 n+3 k+1)}{4}$ for any integers $n$ and $k$. Hence, $|x+y|=\left|\frac{3 \varphi+\pi}{8}-\frac{\pi m}{4}\right|$, where $m-$ is any integer. Since $\frac{\pi}{4}<\varphi<\frac{\pi}{3}$, then $\frac{\pi}{8}<\frac{3 \varphi+\pi}{8}<\frac{\pi}{4}$ and therefore the minimum of the expression $|x+y|$ is achieved at $m=1$.

Answers in different versions: $\frac{\pi}{8}-\frac{3}{8} \arccos \frac{3}{5} ; \frac{\pi}{6}-\frac{1}{3} \arccos \frac{4}{5} ; \frac{3}{8} \arccos \frac{3}{5}-\frac{\pi}{16}$; $\frac{\pi}{3}-\frac{1}{3} \arccos \frac{4}{5}$