\begin{enumerate}
\item $\begin{aligned}[t]
\cos x=\dfrac{1}{2} & \Longleftrightarrow \cos x=\cos \dfrac{\pi}{3}\\
& \Longleftrightarrow x=\dfrac{\pi}{3}+2k\pi\text{ ou } x=-\dfrac{\pi}{3}+2k\pi,\qquad k\in\Z
\end{aligned}$

\newpage

Ainsi, 
\begin{itemize}
\item sur \R, $\boxed{\mathscr{S}=\left\{\dfrac{\pi}{3}+2k\pi~;~-\dfrac{\pi}{3}+2k\pi,\ k\in\Z\right\}}$
\item sur \intervOF{-\pi}{\pi}, $\boxed{\mathscr{S}=\left\{-\dfrac{\pi}{3}~;~\dfrac{\pi}{3} \right\}}$
\end{itemize}

\medskip

\item $\begin{aligned}[t]
\sin x=\dfrac{1}{2} & \Longleftrightarrow \sin x=\sin \dfrac{\pi}{6}\\
& \Longleftrightarrow x=\dfrac{\pi}{6}+2k\pi\text{ ou }x=\pi-\dfrac{\pi}{6}+2k\pi=\dfrac{5\pi}{6}+2k\pi,\quad k\in\Z
\end{aligned}$

Ainsi, 
\begin{itemize}
\item sur \R, $\boxed{\mathscr{S}=\left\{\dfrac{\pi}{6}+2k\pi~;~\dfrac{5\pi}{6}+2k\pi,\ k\in\Z\right\}}$
\item sur \intervOF{-\pi}{\pi}, $\boxed{\mathscr{S}=\left\{\dfrac{\pi}{6}~;~\dfrac{5\pi}{6} \right\}}$
\end{itemize}

\medskip

\item $\begin{aligned}[t]
\cos(2x)=-\dfrac{\sqrt{2}}{2} & \Longleftrightarrow \cos(2x)=\cos\dfrac{3\pi}{4}\\
& \Longleftrightarrow 2x=\dfrac{3\pi}{4}+2k\pi\text{ ou }2x=-\dfrac{3\pi}{4}+2k\pi\\
& \Longleftrightarrow x=\dfrac{3\pi}{8}+k\pi\text{ ou }x=-\dfrac{3\pi}{8}+k\pi
\end{aligned}$

Ainsi,
\begin{itemize}
\item sur \R, $\boxed{\mathscr{S}=\left\{\dfrac{3\pi}{8}+k\pi~;~-\dfrac{3\pi}{8}+k\pi,\ k\in\Z \right\}}$
\item sur \intervOF{-\pi}{\pi}, $\boxed{\mathscr{S}=\left\{-\dfrac{5\pi}{8}~;~-\dfrac{3\pi}{8}~;~\dfrac{3\pi}{8}~;~\dfrac{5\pi}{8} \right\}}$
\end{itemize}

\medskip

\item $\begin{aligned}[t]
\sin\left(3x+\dfrac{\pi}{3}\right)=\dfrac{\sqrt{2}}{2} & \Longleftrightarrow \sin\left(3x+\dfrac{\pi}{3}\right)=\sin\dfrac{\pi}{4}\\
& \Longleftrightarrow 3x+\dfrac{\pi}{3}=\dfrac{\pi}{4}+2k\pi\text{ ou }3x+\dfrac{\pi}{4}=\pi-\dfrac{\pi}{4}+2k\pi,\ k\in\Z\\
& \Longleftrightarrow 3x=\dfrac{\pi}{4}-\dfrac{\pi}{3}+2k\pi\text{ ou }3x=\dfrac{3\pi}{4}-\dfrac{\pi}{3}+2k\pi,\ k\in\Z\\
& \Longleftrightarrow 3x=-\dfrac{\pi}{12}+2k\pi\text{ ou }3x=\dfrac{5\pi}{12}+2k\pi,\ k\in\Z\\
& \Longleftrightarrow x=-\dfrac{\pi}{36}+\dfrac{2k\pi}{3}\text{ ou }x=\dfrac{5\pi}{36}+\dfrac{2k\pi}{3},\ k\in\Z
\end{aligned}$

Ainsi, 
\begin{itemize}
\item sur \R, $\boxed{\mathscr{S}=\left\{ \textcolor{red}{-\dfrac{\pi}{36}+\dfrac{2k\pi}{3}}~;~\textcolor{blue}{\dfrac{5\pi}{36}+\dfrac{2k\pi}{3}},\ k\in\Z \right\}}$
\item sur \intervOF{-\pi}{\pi}, $\boxed{\mathscr{S}=\left\{ \textcolor{red}{-\dfrac{\pi}{36}~;~\dfrac{23\pi}{36}~;~-\dfrac{25\pi}{36}}~;~\textcolor{blue}{\dfrac{5\pi}{36}~;~\dfrac{29\pi}{36}~;~-\dfrac{19\pi}{36}} \right\}}$
\end{itemize}

\paragraph*{Remarque :} pour obtenir ces 6 valeurs, nous avons pris $k=0$, $k=1$ et $k=-1$ pour chaque expression dépendant de $k$.

\medskip

\item $\begin{aligned}[t]
\cos\left(5x-\dfrac{\pi}{6}\right)=-\dfrac{1}{2} & \Longleftrightarrow \cos\left(5x-\dfrac{\pi}{6}\right)=\dfrac{2\pi}{3}\\
& \Longleftrightarrow 5x-\dfrac{\pi}{6}=\dfrac{2\pi}{3}+2k\pi\text{ ou }5x-\dfrac{\pi}{6}=-\dfrac{2\pi}{3}+2k\pi,\ k\in\Z\\
& \Longleftrightarrow 5x=\dfrac{2\pi}{3}+\dfrac{\pi}{6}+2k\pi\text{ ou }5x=-\dfrac{2\pi}{3}+\dfrac{\pi}{6}+2k\pi,\ k\in\Z\\
& \Longleftrightarrow 5x=\dfrac{5\pi}{6}+2k\pi\text{ ou }5x=-\dfrac{3\pi}{6}+2k\pi,\ k\in\Z\\
& \Longleftrightarrow x=\dfrac{\pi}{6}+\dfrac{2k\pi}{5}\text{ ou }x=-\dfrac{\pi}{10}+\dfrac{2k\pi}{5},\ k\in\Z
\end{aligned}$

Ainsi, 
\begin{itemize}
\item sur \R, $\boxed{\mathscr{S}=\left\{ \dfrac{\pi}{6}+\dfrac{2k\pi}{5}~;~-\dfrac{\pi}{10}+\dfrac{2k\pi}{5},\ k\in\Z\right\}}$
\item sur \intervOF{-\pi}{\pi}, 

$\boxed{\mathscr{S}=\left\{\dfrac{\pi}{6}~;~\dfrac{17\pi}{30}~;~\dfrac{29\pi}{30}~;~-\dfrac{7\pi}{30}~;~-\dfrac{19\pi}{30}~;~-\dfrac{\pi}{10}~;~\dfrac{3\pi}{10}~;~\dfrac{7\pi}{10}~;~-\dfrac{\pi}{2}~;~-\dfrac{9\pi}{10} \right\}}$
\end{itemize}
\end{enumerate}