
\begin{enumerate}
\item $\begin{aligned}[t]
\cos(x)=-\dfrac{\sqrt3}{2} & \iff \cos(x)=-\cos\dfrac{\pi}{6}\\
& \iff \cos(x) = \cos\left(\pi-\dfrac{\pi}{6}\right)\\
& \iff \cos(x) = \cos\left(\dfrac{5\pi}{6}\right)\\
& \iff x = \dfrac{5\pi}{6} \text{ ou }x=-\dfrac{5\pi}{6} \quad \text{sur }]-\pi;\pi].
\end{aligned}$

\item $\begin{aligned}[t]
\sin(x)=\dfrac{\sqrt2}{2} & \iff \sin(x)=\sin\left(\dfrac{\pi}{4}\right)\\
& \iff x = \dfrac{\pi}{4}\text{ ou }x = \dfrac{3\pi}{4}\quad\text{sur }]-\pi;\pi].
\end{aligned}$

%\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}
