
\begin{enumerate}
\item $\begin{aligned}[t]
|x - \pi| = 2\pi & \iff x-\pi = 2\pi \qquad\text{ou}\qquad -(x-\pi)=2\pi\\
& \iff x=2\pi+\pi \qquad\text{ou}\qquad x-\pi=-2\pi\\
& \iff x = 3\pi \qquad\text{ou}\qquad x=-\pi.\qquad\qquad\qquad\qquad\qquad \text{ Ainsi, }\boxed{S = \{-\pi;3\pi \}}
\end{aligned}$

\item $\begin{aligned}[t]
|x - 4| = \pi & \iff x-4=\pi \qquad\text{ou}\qquad -(x-4)=\pi\\
& \iff x=\pi+4\qquad\text{ou}\qquad x-4=-\pi\\
& \iff x=\pi+4 \qquad\text{ou}\qquad x = -\pi+4
\end{aligned}$

Donc $\boxed{S = \{4-\pi ; 4+\pi\} }$

\item $\begin{aligned}[t]
|x + 1| > 2\pi & \iff x+1>2\pi \qquad\text{ou}\qquad -(x+1)>2\pi\\
& \iff x>2\pi-1 \qquad\text{ou}\qquad x+1 < -2\pi\\
& \iff x>2\pi-1 \qquad\text{ou}\qquad x < -2\pi-1
\end{aligned}$

Ainsi, $\boxed{S = \intervOO{-\infty}{-2\pi-1} \cup \intervOO{2\pi-1}{+\infty}}$
\end{enumerate}
