\begin{enumerate}
\item $2\cos^2x-3\cos x+1\geqslant0$.

D'après l'exercice précédent, le polynôme $2X^2-3X+1$ se factorise sous la forme :
\[
2(X-1)\left(X-\frac{1}{2}\right).
\]
Ainsi,
\[
2\cos^2x-3\cos x+1=2(\cos x-1)\left(\cos x-\frac{1}{2}\right).
\]
\begin{minipage}{\dimexpr\textwidth-8cm}
\begin{itemize}
\item $-1\leqslant\cos x\leqslant1$ donc $-2\leqslant \cos x-1\leqslant 0$
\item $\begin{aligned}[t]
\cos x-\dfrac{1}{2}>0 & \iff \cos x > \dfrac{1}{2}\\
& \iff x\in\intervFF{-\dfrac{\pi}{3}}{\dfrac{\pi}{3}}
\end{aligned}$
\end{itemize}
\end{minipage}
\begin{minipage}{39mm}
\begin{tikzpicture}
\draw[->,>=latex] (-2,0) -- (2,0) node[below] {cos};
\draw[->,>=latex] (0,-2) -- (0,2) node[right] {sin};
\draw (0,0) circle (1.5cm);
\node[below left] at (0,0) {O};
\draw[very thick,violet] (-60:1.5) arc(-60:60:1.5cm);
\draw[dotted] (-60:1.5cm) -- (60:1.5cm) node[midway,below right] {$\frac{1}{2}$};
\draw[very thick,violet] (0.75,0) -- (1.5,0);
\end{tikzpicture}
\end{minipage}

D'où le tableau suivant :

\begin{center}
\begin{tikzpicture}
\tkzTabInit[lgt=4,color,colorC=violet!20,colorL=violet!20,colorV=violet!20]{$x$/0.8,$\cos x-1$/0.5,$\cos x-\frac{1}{2}$/0.8,$2\cos^2x-3\cos x+1$/0.5}{$-\pi$,$-\frac{\pi}{3}$,$\frac{\pi}{3}$,$\pi$}
\tkzTabLine{,-,,-,,-,0}
\tkzTabLine{,-,0,+,0,-}
\tkzTabLine{,+,0,-,0,+,0}
\end{tikzpicture}
\end{center}

L'ensemble solution de l'inéquation est donc :
\[
\boxed{\mathcal{S}=\intervFF{-\pi}{-\dfrac{\pi}{3}}\cup\intervFF{\frac{\pi}{3}}{\pi}}
\]

\item $2\sin^2x+5\sin x+2<0$.

D'après l'exercice précédent, 
\[
2\sin^2x+5\sin x+2=2(\sin x+2)\left(\sin x+\frac{1}{2}\right).
\]
\begin{minipage}{\dimexpr\textwidth-8cm}
\begin{itemize}
\item $-1\leqslant\sin x\leqslant1$ donc $1\leqslant \sin x+2\leqslant 3$
\item $\begin{aligned}[t]
\sin x+\dfrac{1}{2}<0 & \iff \sin x < -\dfrac{1}{2}\\
& \iff x\in\intervFF{-\dfrac{5\pi}{6}}{-\dfrac{\pi}{6}}
\end{aligned}$
\end{itemize}
\end{minipage}
\begin{minipage}{39mm}
\begin{tikzpicture}
\draw[->,>=latex] (-2,0) -- (2,0) node[below] {cos};
\draw[->,>=latex] (0,-2) -- (0,2) node[right] {sin};
\draw (0,0) circle (1.5cm);
\node[below left] at (0,0) {O};
\draw[very thick,violet] (-150:1.5) node[below left] {$-\frac{5\pi}{6}$} arc(-150:-30:1.5cm) node[below right] {$-\frac{\pi}{6}$};
\draw[dotted] (-30:1.5cm) -- (210:1.5cm) node[midway,above right] {$-\frac{1}{2}$};
\draw[very thick,violet] (0,-0.75) -- (0,-1.5);
\end{tikzpicture}
\end{minipage}

D'où le tableau suivant :

\begin{center}
\begin{tikzpicture}
\tkzTabInit[lgt=4,color,colorC=violet!20,colorL=violet!20,colorV=violet!20]{$x$/0.8,$\sin x+2$/0.5,$\sin x+\frac{1}{2}$/0.8,$2\sin^2x+5\sin x+2$/0.5}{$-\pi$,$-\frac{5\pi}{6}$,$-\frac{\pi}{6}$,$\pi$}
\tkzTabLine{,+,,+,,+,}
\tkzTabLine{,+,0,-,0,+}
\tkzTabLine{,+,0,-,0,+,}
\end{tikzpicture}
\end{center}

L'ensemble solution de l'inéquation est donc :
\[
\boxed{\mathcal{S}=\intervOO{-\dfrac{5\pi}{6}}{-\dfrac{\pi}{6}}}
\]

\item $3\cos^2x-\left(9+\dfrac{3}{\sqrt{2}}\right)\cos x+\dfrac{9}{\sqrt{2}}\leqslant0$.

D'après l'exercice précédent, 
\[
P(x)=3\cos^2x-\left(9+\dfrac{3}{\sqrt{2}}\right)\cos x+\dfrac{9}{\sqrt{2}}=3\left(\cos x-\dfrac{\sqrt{2}}{2}\right)(\cos x-3).
\]
\begin{minipage}{\dimexpr\textwidth-8cm}
\begin{itemize}
\item $-1\leqslant\cos x\leqslant1$ donc $-4\leqslant \cos x-3\leqslant -2$
\item $\begin{aligned}[t]
\cos x-\dfrac{\sqrt{2}}{2}>0 
& \iff \cos x > \dfrac{\sqrt{2}}{2}\\
& \iff x\in\intervFF{-\dfrac{\pi}{4}}{\dfrac{\pi}{4}}
\end{aligned}$
\end{itemize}
\end{minipage}
\begin{minipage}{39mm}
\begin{tikzpicture}
\draw[->,>=latex] (-2,0) -- (2,0) node[below] {cos};
\draw[->,>=latex] (0,-2) -- (0,2) node[right] {sin};
\draw (0,0) circle (1.5cm);
\node[below left] at (0,0) {O};
\draw[very thick,violet] (-45:1.5) node[below right] {$-\frac{\pi}{4}$} arc(-45:45:1.5cm) node[above right] {$\frac{\pi}{4}$};
\draw[dotted] (-45:1.5cm) -- (45:1.5cm) node[midway,below left] {$\frac{\sqrt{2}}{2}$};
\end{tikzpicture}
\end{minipage}

D'où le tableau suivant :

\begin{center}
\begin{tikzpicture}
\tkzTabInit[lgt=4,color,colorC=violet!20,colorL=violet!20,colorV=violet!20]{$x$/0.8,$\cos x-3$/0.5,$\cos x-\frac{\sqrt{2}}{2}$/0.8,$P(x)$/0.5}{$-\pi$,$-\frac{\pi}{4}$,$\frac{\pi}{4}$,$\pi$}
\tkzTabLine{,-,,-,,-,}
\tkzTabLine{,-,0,+,0,-}
\tkzTabLine{,+,0,-,0,+,}
\end{tikzpicture}
\end{center}

L'ensemble solution de l'inéquation est donc :
\[
\boxed{\mathcal{S}=\intervFF{-\dfrac{\pi}{4}}{\dfrac{\pi}{4}}}
\]
\end{enumerate}
\newpage