
\begin{align*}
4\cos^2x-3=0 & \iff 4\cos^2x=3\\
& \iff \cos^2x=\dfrac{3}{4}\\
& \iff \cos x = -\sqrt{\dfrac{3}{4}}=-\dfrac{\sqrt3}{2}\text{  ou  }\cos x = \dfrac{\sqrt3}{2}. 
\end{align*}
\newpage
On cherche donc tous les angles en radians sur \intervOF{-\pi}{\pi}  tels que $\cos x = -\sqrt{\dfrac{3}{4}}=-\dfrac{\sqrt3}{2}\text{  ou  }\cos x = \dfrac{\sqrt3}{2}$.

On trouve alors :
\begin{itemize}
\item $x=-\dfrac{5\pi}{6}$ et $x=\dfrac{5\pi}{6}$ pour $\cos x = -\dfrac{\sqrt3}{2}$,
\item $x=-\dfrac{\pi}{6}$ et $x=\dfrac{\pi}{6}$ pour $\cos x = \dfrac{\sqrt3}{2}$.
\end{itemize}

\begin{center}
\begin{tikzpicture}[scale=3]
\draw[gray!30] (-1,-1) grid[step=0.1] (1,1);
\draw[->,>=latex] (-1.1,0) -- (1.1,0) node[below right] {cos};
\draw[->,>=latex] (0,-1.1) -- (0,1.1) node[right] {sin};
\draw (0,0) circle (1);
\fill (-1,0) circle (0.66pt) node[above left] {$-\pi\longleftrightarrow\pi$};
\fill[red] (-150:1) circle (0.66pt) node[below left] {$-\frac{5\pi}{6}$};
\fill[red] (-30:1) circle (0.66pt) node[below right] {$-\frac{\pi}{6}$};
\fill[red] (30:1) circle (0.66pt) node[above right] {$\frac{\pi}{6}$};
\fill[red] (150:1) circle (0.66pt) node[above left] {$\frac{5\pi}{6}$};
\draw[dashed,thick,red] (150:1) -- (-150:1);
\draw[dashed,thick,red] (30:1) -- (-30:1);
\fill[red] ({-sqrt(3)/2},0) circle (0.66pt) node[above right] {$-\frac{\sqrt3}{2}$};
\fill[red] ({sqrt(3)/2},0) circle (0.66pt) node[above left] {$\frac{\sqrt3}{2}$};
\end{tikzpicture}
\end{center}