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**woo** From $\cos{x}(14\sin{x} - 2\cos{x}) = 0$, we get $\cos{x}=0$ or $14\sin{x}-2\cos{x}=0$. From equation $14\sin{x}-2\cos{x}=0$, by assuming that $sin{x}\neq{0}$, we get the solution $x=\tan^{-1}(1/7)$. However, for the equation $14\sin{x}-2\cos{x}=0$, we have not yet rule out the possibility that $sin{x}=0$ since we previously assume that $sin{x}\neq{0}$. How can we show that x for which $\sin{x}=0$ is not a solution of $14\sin{x}-2\cos{x}=0$ and hence not a solution of the original equation?