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

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

S = \left\{ -\dfrac{2\pi}{3}+k2\pi, k\in\mathbb{Z}; \dfrac{2\pi}{3}+k2\pi , k\in\mathbb{Z}\right\}

S = \left\{-\dfrac{\pi}{3}+k2\pi, k\in\mathbb{Z}; \dfrac{\pi}{3}+k2\pi , k\in\mathbb{Z}\right\}

S = \left\{ -\dfrac{4\pi}{3}+k2\pi, k\in\mathbb{Z}; \dfrac{4\pi}{3}+k2\pi , k\in\mathbb{Z}\right\}

S =\varnothing

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

S = \left\{ -\dfrac{\pi}{4}+k2\pi, k\in\mathbb{Z}; \dfrac{\pi}{4}+k2\pi, k\in\mathbb{Z} \right\}

S = \left\{ -\dfrac{3\pi}{4}+k2\pi, k\in\mathbb{Z}; \dfrac{3\pi}{4}+k2\pi, k\in\mathbb{Z} \right\}

S = \left\{ -\dfrac{\pi}{2}+k2\pi, k\in\mathbb{Z}; \dfrac{\pi}{2}+k2\pi, k\in\mathbb{Z} \right\}

S = \varnothing

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

S = \left\{ -\dfrac{\pi}{6}+k2\pi, k\in\mathbb{Z}; \dfrac{\pi}{6}+k2\pi, k\in\mathbb{Z} \right\}

S = \left\{ -\dfrac{3\pi}{6}+k2\pi, k\in\mathbb{Z}; \dfrac{3\pi}{6}+k2\pi, k\in\mathbb{Z} \right\}

S = \left\{ -\dfrac{\pi}{2}+k2\pi, k\in\mathbb{Z}; \dfrac{\pi}{2}+k2\pi, k\in\mathbb{Z} \right\}

S = \varnothing

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

S = \left\{ -\dfrac{\pi}{3}+k2\pi, k\in\mathbb{Z}; \dfrac{\pi}{3}+k2\pi, k\in\mathbb{Z} \right\}

S = \left\{ -\dfrac{3\pi}{4}+k2\pi, k\in\mathbb{Z}; \dfrac{3\pi}{4}+k2\pi, k\in\mathbb{Z} \right\}

S = \left\{ -\dfrac{\pi}{2}+k2\pi, k\in\mathbb{Z}; \dfrac{\pi}{2}+k2\pi, k\in\mathbb{Z} \right\}

S = \varnothing

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

S = \left\{ -\dfrac{5\pi}{6}+k2\pi, k\in\mathbb{Z}; \dfrac{5\pi}{6}+k2\pi, k\in\mathbb{Z} \right\}

S = \left\{ -\dfrac{3\pi}{6}+k2\pi, k\in\mathbb{Z}; \dfrac{3\pi}{6}+k2\pi, k\in\mathbb{Z} \right\}

S = \left\{ -\dfrac{\pi}{2}+k2\pi, k\in\mathbb{Z}; \dfrac{\pi}{2}+k2\pi, k\in\mathbb{Z} \right\}

S = \varnothing