\begin{methode}
Nous allons utiliser ici les équivalences suivantes, où $k\in\Z$:
\begin{align*}
\cos x = \cos a & \iff x = a + 2k\pi \text{ ou } x = -a+2k\pi.\\
\sin x = \sin a & \iff x = a + 2k\pi\text{ ou } x = \pi-a+2k\pi
\end{align*}
\end{methode}

\begin{questions}
\item $\begin{aligned}[t]
\cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{2} &
\iff \cos\left(x+\dfrac{\pi}{3}\right)=\cos\dfrac{\pi}{3}\\
& \iff \begin{cases}
x+\dfrac{\pi}{3} = \dfrac{\pi}{3}+2k\pi\\[8pt]
x+\dfrac{\pi}{3} = -\dfrac{\pi}{3}+2k\pi
\end{cases},\ k\in\Z\\
& \iff \begin{cases}
x=\dfrac{\pi}{3}-\dfrac{\pi}{3}+2k\pi\\[8pt]
x = -\dfrac{\pi}{3}-\dfrac{\pi}{3}+2k\pi
\end{cases},\quad k\in\Z\\
& \iff \begin{cases}
x = 0+2k\pi\\[8pt]
x = -\dfrac{2\pi}{3}+2k\pi
\end{cases},\quad k\in\Z
\end{aligned}$

\Par

L'ensemble des solutions de l'équation $\cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}$ sur $]-\pi;\pi]$ est donc:

\Par

\[
\mathscr{S}=\left\{ 0 ; -\dfrac{2\pi}{3} \right\}.
\]

\Par

\item $\begin{aligned}[t]
\sin\left(2x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2} & \iff \sin\left(2x-\dfrac{\pi}{4}\right)=\sin\left(-\dfrac{\pi}{6}\right)\\
& \iff \begin{cases}
2x-\dfrac{\pi}{4} & = -\dfrac{\pi}{6} + 2k\pi\\[8pt]
2x-\dfrac{\pi}{4} & = \pi - \left(-\dfrac{\pi}{6}\right) + 2k\pi
\end{cases},\qquad k\in\Z\\
& \iff \begin{cases}
2x & = -\dfrac{\pi}{6}+\dfrac{\pi}{4} + 2k\pi\\[8pt]
2x & = \pi + \dfrac{\pi}{6} + \dfrac{\pi}{4} + 2k\pi
\end{cases},\qquad k\in\Z\\
& \iff \begin{cases}
2x & = \dfrac{\pi}{12} + 2k\pi\\[8pt]
2x & = \dfrac{17\pi}{12} + 2k\pi
\end{cases},\qquad k\in\Z\\
& \iff \begin{cases}
x & = \dfrac{\pi}{24} + k\pi\\[8pt]
x & = \dfrac{17\pi}{24} + k\pi
\end{cases},\qquad k\in\Z
\end{aligned}$

\Par

Ainsi, les solutions de l'équation sur $]-\pi;\pi]$ sont obtenues en prenant $k=0$ et $k=-1$; on obtient alors l'ensemble des solutions suivant:

\Par

\[
\mathscr{S}=\left\{ -\dfrac{23\pi}{24} ~;~ -\dfrac{7\pi}{24} ~;~ \dfrac{\pi}{24} ~;~ \dfrac{17\pi}{24} \right\}.
\]
\end{questions}
