Résoudre une équation trigonométrique du type sin(x)=y - Tle - Exercice Mathématiques

\sin\left(x\right) =-\dfrac{1}{2}

S = \left\{ -\dfrac{\pi}{6}+k2\pi; \dfrac{7\pi}{6}+k2\pi \right\}

S = \left\{ \dfrac{\pi}{6}+k2\pi; \dfrac{7\pi}{6}+k2\pi \right\}

S = \left\{ -\dfrac{\pi}{6}+k2\pi; -\dfrac{7\pi}{6}+k2\pi \right\}

S = \left\{\dfrac{\pi}{6}+k2\pi; -\dfrac{7\pi}{6}+k2\pi \right\}

\sin\left(x\right) =-\dfrac{\sqrt3}{2}

S = \left\{ -\dfrac{\pi}{3}+k2\pi; \dfrac{4\pi}{3}+k2\pi \right\}

S = \left\{ \dfrac{\pi}{3}+k2\pi; -\dfrac{4\pi}{3}+k2\pi \right\}

S = \left\{ -\dfrac{\pi}{6}+k2\pi; -\dfrac{7\pi}{6}+k2\pi \right\}

S = \left\{\dfrac{\pi}{6}+k2\pi; -\dfrac{7\pi}{6}+k2\pi \right\}

\sin\left(x\right) = \dfrac{\sqrt2}{2}

S = \left\{ \dfrac{\pi}{4}+k2\pi; \dfrac{3\pi}{4}+k2\pi \right\}

S = \left\{ - \dfrac{\pi}{4}+k2\pi; \dfrac{3\pi}{4}+k2\pi \right\}

S = \left\{ \dfrac{\pi}{4}+k2\pi; -\dfrac{3\pi}{4}+k2\pi \right\}

S = \left\{ - \dfrac{\pi}{4}+k2\pi; - \dfrac{3\pi}{4}+k2\pi \right\}

\sin\left(x\right) = -\dfrac{\sqrt2}{2}

S = \left\{ -\dfrac{\pi}{4}+k2\pi; \dfrac{5\pi}{4}+k2\pi \right\}

S = \left\{ \dfrac{\pi}{4}+k2\pi; \dfrac{5\pi}{4}+k2\pi \right\}

S = \left\{ -\dfrac{\pi}{4}+k2\pi; -\dfrac{5\pi}{4}+k2\pi \right\}

S = \left\{\dfrac{\pi}{4}+k2\pi; -\dfrac{5\pi}{4}+k2\pi \right\}

\sin\left(x\right) =0

S = \left\{ \pi +k\pi \right\}

S = \left\{ k\pi \right\}

S = \left\{ -\dfrac \pi 2+k\pi \right\}

S = \left\{ -\dfrac \pi 3+k\pi \right\}