\require{AMSmath}

Een goniometrische vergelijking oplossen

$ \eqalign{ & \cos (x) = \sin \left( {{1 \over 2}x} \right) \cr & Formulekaart: \cr & \cos 2A = 1 - 2\sin ^2 A \cr & of\,\,\,ook: \cr & \cos (x) = 1 - 2\sin ^2 \left( {{1 \over 2}x} \right) \cr & Invullen\,\,\,geeft: \cr & 1 - 2\sin ^2 \left( {{1 \over 2}x} \right) = \sin \left( {{1 \over 2}x} \right) \cr & 2\sin ^2 \left( {{1 \over 2}x} \right) + \sin \left( {{1 \over 2}x} \right) - 1 = 0 \cr & Kies\,\,\,y = \sin \left( {{1 \over 2}x} \right) \cr & 2y^2 + y - 1 = 0 \cr & (2y - 1)(y + 1) = 0 \cr & 2y - 1 = 0 \vee y + 1 = 0 \cr & y = {1 \over 2} \vee y = - 1 \cr & \sin \left( {{1 \over 2}x} \right) = {1 \over 2} \vee \sin \left( {{1 \over 2}x} \right) = - 1 \cr } $

$ \eqalign{ & {1 \over 2}x = {1 \over 6}\pi + k \cdot 2\pi \vee {1 \over 2}x = {5 \over 6}\pi + k \cdot 2\pi \vee {1 \over 2}x = 1{1 \over 2}\pi + k \cdot 2\pi \cr & x = {1 \over 3}\pi + k \cdot 4\pi \vee x = 1{2 \over 3}\pi + k \cdot 4\pi \vee x = 3\pi + k \cdot 4\pi \cr} $


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