On sait que $x\in]-\pi;0]$. De plus,
\begin{align*}
\cos^2x+\sin^2x = 1 & \iff \sin^2x = 1 - \left(\frac{1+\sqrt5}{4}\right)^2\\
& \iff \sin^2x = 1 - \frac{1+2\sqrt5+5}{16}\\
& \iff \sin^2x = \frac{16-6-2\sqrt5}{16}\\
& \iff \sin^2x = \frac{10-2\sqrt5}{16}\\
& \iff \sin x = -\sqrt{\frac{10-2\sqrt5}{16}}
\intertext{(car $x\in]-\pi;0]\iff\sin x \leqslant 0$)}
% ERREUR % & \iff \sin x = -\frac{\sqrt{10-2\sqrt5}}{16}.
 & \iff \sin x = -\frac{\sqrt{10-2\sqrt5}}{4}.
\end{align*}
