
\begin{enumerate}
\item $\begin{aligned}[t]
\cos(x)=-\dfrac{\sqrt3}{2} & \iff \cos(x)=-\cos\dfrac{\pi}{6}\\
& \iff \cos(x) = \cos\left(\pi-\dfrac{\pi}{6}\right)\\
& \iff \cos(x) = \cos\left(\dfrac{5\pi}{6}\right)\\
& \iff x = \dfrac{5\pi}{6}+2k\pi\text{ ou }x=-\dfrac{5\pi}{6}+2k\pi,\ k\in\Z.
\end{aligned}$

Ainsi, sur $]-\pi;\pi]$, les solutions sont : $-\dfrac{5\pi}{6}$ et $\dfrac{5\pi}{6}$.

\item $\begin{aligned}[t]
\sin(x)=\dfrac{\sqrt2}{2} & \iff \sin(x)=\sin\left(\dfrac{\pi}{4}\right)\\
& \iff x = \dfrac{\pi}{4}+2k\pi\text{ ou }x = \pi-\dfrac{\pi}{4}+2k\pi,\ k\in\Z\\
& \iff x = \dfrac{\pi}{4}+2k\pi\text{ ou }x = \dfrac{3\pi}{4}+2k\pi,\ k\in\Z.\\
\end{aligned}$

Ainsi, sur $]-\pi;\pi]$, les solutions sont :$\dfrac{\pi}{4}$ et $\dfrac{3\pi}{4}$.

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

Ainsi, sur $]-\pi;\pi]$, les solutions sont (en prenant $k=0$ et $k=1)$ : 0, $\pi$, $-\dfrac{\pi}{3}$ et $\dfrac{2\pi}{3}$.

\item Pour résoudre l'inéquation $\sin\left(3x-\dfrac{\pi}{6}\right)\geqslant\dfrac{1}{2}$, on peut poser $X=3x-\dfrac{\pi}{6}$ dans un premier temps.
\[
\sin X \geqslant \dfrac{1}{2} \iff \sin X \geqslant \sin \dfrac{\pi}{6} \iff  \dfrac{\pi}{6} \leqslant X \leqslant \pi-\dfrac{\pi}{6}
\]
\begin{center}
\begin{tikzpicture}
\draw (0,0) circle (2cm);
\draw[->,>=latex] (-2.2,0) -- (2.2,0);
\draw[->,>=latex] (0,-2.2) -- (0,2.2);
\draw[very thick,red] (30:2) arc(30:150:2);
\draw[dashed] (0,0) -- (30:2)-- (150:2) -- cycle;
\end{tikzpicture}
\end{center}
\begin{align*}
\sin\left(3x-\dfrac{\pi}{6}\right)\geqslant\dfrac{1}{2} & \iff \dfrac{\pi}{6}+2k\pi \leqslant 3x-\dfrac{\pi}{6} \leqslant\pi-\dfrac{\pi}{6} +2k\pi,\ k\in\Z\\
& \iff \dfrac{\pi}{3}+2k\pi \leqslant 3x \leqslant \pi+2k\pi,\ k\in\Z\\
& \iff \dfrac{\pi}{9}+\dfrac{2k\pi}{3} \leqslant x \leqslant \dfrac{\pi}{3} +\dfrac{2k\pi}{3},\ k\in\Z
\end{align*}
\begin{itemize}
\item Pour $k=0$, on obtient : $\dfrac{\pi}{9} \leqslant x \leqslant \dfrac{\pi}{3}$.
\item Pour $k=-1$, on obtient : $-\dfrac{5\pi}{9} \leqslant x \leqslant -\dfrac{\pi}{3}$.
\item Pour $k=1$, on obtient : $\dfrac{7\pi}{9} \leqslant x \leqslant \pi$.
\end{itemize}
\begin{center}
\begin{tikzpicture}
\draw (0,0) circle (2cm);
\draw[->,>=latex] (-2.2,0) -- (2.2,0);
\draw[->,>=latex] (0,-2.2) -- (0,2.2);
\draw[very thick,red] (20:2) arc(20:60:2);
\node[right,red] at (20:2) {$\frac{\pi}{9}$};
\node[above right,red] at (60:2) {$\frac{\pi}{3}$};
%
\draw[very thick,red] (-100:2) arc(-100:-60:2);
\node[below,red] at (-60:2) {$-\frac{\pi}{3}$};
\node[below left,red] at (-100:2) {$-\frac{5\pi}{9}$};
%
\draw[very thick,red] (140:2) arc(140:180:2);
\node[above,red] at (140:2) {$\frac{7\pi}{9}$};
\node[above left,red] at (180:2) {$\pi$};
\end{tikzpicture}
\end{center}
\end{enumerate}
