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Convertissez en radians l'angle \(-160^\circ \).
\( -160\mbox{ radians}\)
\( \dfrac{8\pi}{9} \mbox{ radians}\)
\(\dfrac{10\pi}{9} \mbox{ radians}\)
\( -\dfrac{10\pi}{9}\mbox{ radians}\)
Résolvez l'équation \(\cos x = \dfrac{\sqrt{3}}{2} \).
\(S=\left\{\dfrac{\pi}{6}\right\} \)
\( S=\left\{\dfrac{\pi}{6},\, \dfrac{-\pi}{6}\right\} \)
\( S=\left\{\dfrac{\pi}{3}+2k\pi,\, \dfrac{-\pi}{3}+2k\pi;\, k\in\mathbb{Z}\right\} \)
\( S=\left\{\dfrac{\pi}{6}+2k\pi,\, \dfrac{-\pi}{6}+2k\pi;\, k\in\mathbb{Z}\right\} \)
Convertissez en radians l'angle \(\normalsize 150^\circ\).
\(\dfrac{5\pi}{6}\mbox{ radians}\)
\(150\mbox{ radians}\)
\(\dfrac{5\pi}{3}\mbox{ radians}\)
\(\dfrac{\pi}{150}\mbox{ radians}\)
Résolvez l'équation \( \cos x = -{1\over 2} \).
\( S=\left\{\dfrac{2\pi}{3}\right\} \)
\( S=\left\{\dfrac{2\pi}{3},\, \dfrac{4\pi}{3}\right\} \)
\( S=\left\{\dfrac{2\pi}{3}+2k\pi,\, \dfrac{4\pi}{3}+2k\pi;\, k\in\mathbb{Z}\right\} \)
\( S=\left\{\dfrac{\pi}{3}+2k\pi,\, -\dfrac{\pi}{3}+2k\pi;\, k\in\mathbb{Z}\right\} \)
\( \sin (-a)= \)
\( \sin a \)
\( -\sin a \)
\(\cos a \)
\( -\cos a \)
Résolvez l'équation \(\sin x = \dfrac{\sqrt{2}}{2} \).
\( S=\left\{\dfrac{\pi}{4}\right\} \)
\( S=\left\{\dfrac{\pi}{4},\, \dfrac{3\pi}{4}\right\} \)
\( S=\left\{\dfrac{\pi}{4}+2k\pi,\, \dfrac{3\pi}{4}+2k\pi;\, k\in\mathbb{Z}\right\} \)
\( S=\left\{\dfrac{\pi}{4}+2k\pi,\, \dfrac{7\pi}{4}+2k\pi;\, k\in\mathbb{Z}\right\} \)
Sachant que \(ABCDEF\) est un hexagone régulier inscrit dans un cercle de centre \(O\) , comparez les angles \(\widehat{CAD}\) et \( \widehat{CFD} \).
\( \widehat{CAD}=\widehat{CFD} \)
\( 2\widehat{CAD}=\widehat{CFD} \)
\(\widehat{CAD}=2\widehat{CFD} \)
\( \widehat{CAD}=\dfrac{1}{2}\widehat{CFD} \)
\(\sin (3\pi +a)= \)
\( \cos a \)
\( \pi+\sin a \)
Convertissez en degrés l'angle \(5\pi\).
\(\pi\mbox{ degrés}\)
\( 5\mbox{ degrés}\)
\(90\mbox{ degrés}\)
\(180\mbox{ degrés}\)
Si \(\alpha=53^{\circ}\) , alors le complémentaire de \(\alpha\) vaut
\( 37^{\circ}\)
\( 127^{\circ} \)
\( 143^{\circ} \)
\( 413^{\circ} \)
