Résoudre une équation trigonométrique du type sin(ax+b)=y - Tle - Exercice Mathématiques

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

S = \left\{ \dfrac{\pi}{6}+k\pi; \dfrac{\pi}{3}+k\pi \right\}

S = \left\{ \dfrac{\pi}{3}+2k\pi \right\}

S = \left\{ \dfrac{\pi}{6}+k\pi; \dfrac{2\pi}{3}+k\pi \right\}

S = \varnothing

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

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

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

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

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

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

S = \left\{ \dfrac{5\pi}{4} +k2\pi ; \dfrac{3\pi}{4} +k2\pi\right\}

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

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

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

\sin\left(2x-\dfrac{\pi}{6}\right) =\dfrac{\sqrt3}{2}

S = \left\{ \dfrac{\pi}{4} +k\pi ; \dfrac{5\pi}{12} +k\pi\right\}

S = \left\{ \dfrac{\pi}{4} +k\pi ; \dfrac{\pi}{12} +k\pi\right\}

S = \left\{ -\dfrac{\pi}{4} +k\pi ; \dfrac{5\pi}{12} +k\pi\right\}

S = \left\{ \dfrac{\pi}{4} +k\pi ; \dfrac{7\pi}{12} +k\pi\right\}

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

S = \left\{ \dfrac{3\pi}{8}+k\pi; \dfrac{\pi}{8}+k\pi \right\}

S = \left\{ -\dfrac{3\pi}{8}+k\pi; \dfrac{\pi}{8}+k\pi \right\}

S = \left\{ \dfrac{3\pi}{8}+k\pi; -\dfrac{\pi}{8}+k\pi \right\}

S = \left\{ \dfrac{3\pi}{8}+2k\pi \right\}