Solve the equation: $\;$ $\sin \left(3x - 2\right) = -1$
Given equation: $\;\;\;$ $\sin \left(3x - 2\right) = -1$
i.e. $\;$ $3x - 2 = \sin^{-1} \left(-1\right)$
i.e. $\;$ $3x - 2 = - \sin^{-1} \left(1\right)$ $\;\;\;$ $\left\{\because \;\; \sin^{-1} \left(-x\right) = - \sin^{-1} x \;\; \forall \; x \in \left[-1, 1\right]\right\}$
i.e. $\;$ $3x - 2 = \dfrac{- \pi}{2} + 2 n \pi \;\;\; n \in Z$
i.e. $\;$ $3x = \dfrac{4 n \pi - \pi}{2} + 2 \;\;\; n \in Z$
i.e. $\;$ $x = \dfrac{4n \pi - \pi}{6} + \dfrac{2}{3} \;\;\; n \in Z$
i.e. $\;$ $x = \dfrac{\pi \left(4n - 1\right) + 4}{6} \;\;\; n \in Z$