Auto-Math
Si \(\sin\theta=\dfrac{3}{5} \) alors \(cotg\,\theta=\)
\( \dfrac{2}{5} \)
\( \dfrac{3}{4} \)
\( \dfrac{4}{3} \)
n'existe pas
Résolvez l'équation \(\sin 2x = \sin \dfrac{\pi}{4}\) .
\( S=\left\{\dfrac{\pi}{8}\right\} \)
\( S=\left\{\dfrac{\pi}{8},\, \dfrac{3\pi}{8}\right\} \)
\(S=\left\{\dfrac{\pi}{8}+2k\pi,\, \dfrac{3\pi}{8}+2k\pi;\, k\in\mathbb{Z}\right\} \)
\( S=\left\{\dfrac{\pi}{8}+k\pi,\, \dfrac{3\pi}{8}+k\pi;\, k\in\mathbb{Z}\right\} \)
Résolvez l'équation \(\cos(3x+\pi) = \cos x \).
\( S=\left\{-\dfrac{\pi}{2}+k\pi,\, -\dfrac{\pi}{4}+k\dfrac{\pi}{2};\, k\in\mathbb{Z}\right\} \)
\( S=\left\{-\dfrac{\pi}{2},\, -\dfrac{\pi}{4};\, k\in\mathbb{Z}\right\} \)
\( S=\left\{-\dfrac{\pi}{2}+k\pi;\, k\in\mathbb{Z}\right\} \)
\( S=\left\{-\dfrac{\pi}{2}+2k\pi,\, -\dfrac{\pi}{4}+2k\dfrac{\pi}{2};\, k\in\mathbb{Z}\right\} \)
Résolvez l'équation \(tg\, 3x = \dfrac{\sqrt{3}}{3}\) .
\( S=\left\{\dfrac{\pi}{18}\right\} \)
\( S=\left\{\dfrac{\pi}{18}+k\dfrac{\pi}{3};\, k\in\mathbb{Z}\right\} \)
\( S=\left\{\dfrac{\pi}{18}+2k\dfrac{\pi}{3};\, k\in\mathbb{Z}\right\} \)
\( S=\left\{\dfrac{\pi}{18}+2k\pi;\, k\in\mathbb{Z}\right\} \)
Déterminez à l'aide du cercle trigonométrique la valeur de \(\cos\dfrac{2\pi}{3} \).
\( \dfrac{1}{2} \)
\( \dfrac{\sqrt{3}}{2} \)
\( -\dfrac{1}{2} \)
\( -\dfrac{\sqrt{3}}{2} \)
Résolvez l'équation \(\cos x = \cos \dfrac{\pi}{3} \).
\( S=\left\{\dfrac{\pi}{3}\right\} \)
\( S=\left\{\dfrac{\pi}{3}+2k\pi,\, -\dfrac{\pi}{3}+2k\pi;\, k\in\mathbb{Z}\right\} \)
\( S=\left\{\dfrac{\pi}{3},\, -\dfrac{\pi}{3}\right\} \)
\( S=\left\{\dfrac{\pi}{3}+2k\pi,\, \dfrac{2\pi}{3}+2k\pi;\, k\in\mathbb{Z}\right\} \)
\(\sin (2\pi -a)= \)
\( \sin a \)
\( -\sin a \)
\(\cos a \)
\(2\pi-\sin a \)
Résolvez l'équation \( \cos x = -{1\over 2} \).
\( S=\left\{\dfrac{2\pi}{3}\right\} \)
\( S=\left\{\dfrac{2\pi}{3},\, \dfrac{4\pi}{3}\right\} \)
\( S=\left\{\dfrac{2\pi}{3}+2k\pi,\, \dfrac{4\pi}{3}+2k\pi;\, k\in\mathbb{Z}\right\} \)
Si \(tg\,\theta=\dfrac{5}{12}\) alors \(cotg\, \theta=\)
\(\dfrac{12}{5} \)
\( \dfrac{5}{12} \)
\( \dfrac{7}{12} \)
Si \(\alpha=53^{\circ}\) , alors l'opposé de \(\alpha\) vaut
\( 35^{\circ} \)
\( 233^{\circ} \)
\( 413^{\circ} \)
\( -53^{\circ} \)
