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Résolvez l'équation \(\cos 2x = \dfrac{\sqrt{2}}{2}\) .
\( S=\left\{\dfrac{\pi}{8}+k\pi,\, -\dfrac{\pi}{8}+k\pi;\, k\in\mathbb{Z}\right\} \)
\( S=\left\{\dfrac{\pi}{8}+2k\pi,\, -\dfrac{\pi}{8}+2k\pi;\, k\in\mathbb{Z}\right\} \)
\( S=\left\{\dfrac{\pi}{4}\right\} \)
\( S=\left\{\dfrac{\pi}{8}\right\} \)
Convertissez en radians l'angle \(390^\circ \).
\(30\mbox{ radians}\)
\(\dfrac{\pi}{3} \mbox{ radians}\)
\( \dfrac{\pi}{6}\mbox{ radians}\)
\( 2\pi \mbox{ radians}\)
Si \(\alpha=53^{\circ}\) , alors le complémentaire de \(\alpha\) vaut
\( 37^{\circ}\)
\( 127^{\circ} \)
\( 143^{\circ} \)
\( 413^{\circ} \)
Résolvez l'équation \(\cos x = \dfrac{\sqrt{3}}{2} \).
\(S=\left\{\dfrac{\pi}{6}\right\} \)
\( S=\left\{\dfrac{\pi}{6},\, \dfrac{-\pi}{6}\right\} \)
\( S=\left\{\dfrac{\pi}{3}+2k\pi,\, \dfrac{-\pi}{3}+2k\pi;\, k\in\mathbb{Z}\right\} \)
\( S=\left\{\dfrac{\pi}{6}+2k\pi,\, \dfrac{-\pi}{6}+2k\pi;\, k\in\mathbb{Z}\right\} \)
Sachant que \(ABCDEF\) est un hexagone régulier inscrit dans un cercle de centre \(O\) , comparez les angles \(\widehat{CAD}\) et \( \widehat{CFD} \).
\( \widehat{CAD}=\widehat{CFD} \)
\( 2\widehat{CAD}=\widehat{CFD} \)
\(\widehat{CAD}=2\widehat{CFD} \)
\( \widehat{CAD}=\dfrac{1}{2}\widehat{CFD} \)
\(\sin (2\pi -a)= \)
\( \sin a \)
\( -\sin a \)
\(\cos a \)
\(2\pi-\sin a \)
Résolvez l'équation \(\sin x = \dfrac{\sqrt{2}}{2} \).
\( S=\left\{\dfrac{\pi}{4},\, \dfrac{3\pi}{4}\right\} \)
\( S=\left\{\dfrac{\pi}{4}+2k\pi,\, \dfrac{3\pi}{4}+2k\pi;\, k\in\mathbb{Z}\right\} \)
\( S=\left\{\dfrac{\pi}{4}+2k\pi,\, \dfrac{7\pi}{4}+2k\pi;\, k\in\mathbb{Z}\right\} \)
Convertissez en radians l'angle \(45^\circ \).
\(45\mbox{ radians}\)
\( \dfrac{\pi}{2}\mbox{ radians}\)
\(\dfrac{\pi}{4}\mbox{ radians}\)
\( \dfrac{\pi}{45} \mbox{ radians}\)
Donnez la valeur de \( \sin {\pi \over 2}\) .
1
-1
0
90
Donnez la valeur de \( cotg\left(\dfrac{2\pi}{3}\right) \).
\(60 \)
\( -\sqrt{3} \)
\( -\dfrac{\sqrt{3}}{3} \)
\( \dfrac{\sqrt{3}}{3} \)
