Find the general solution of $\sin 2 x+\cos x=0$.

The given equation is $\sin 2x + \cos x = 0$.

Using the double angle identity $\sin 2x = 2 \sin x \cos x$, the equation becomes $2 \sin x \cos x + \cos x = 0$.

Factoring out $\cos x$, we get $\cos x (2 \sin x + 1) = 0$.

This implies either $\cos x = 0$ or $2 \sin x + 1 = 0$.

For $\cos x = 0$, the general solution is $x = (2n + 1) \dfrac{\pi}{2}$, which can also be written as $n \pi \pm \dfrac{\pi}{2}$.

For $2 \sin x + 1 = 0$, we have $\sin x = -\dfrac{1}{2}$.

Since $\sin x = -\sin \dfrac{\pi}{6} = \sin(-\dfrac{\pi}{6}) = \sin(\pi + \dfrac{\pi}{6}) = \sin(\dfrac{7\pi}{6})$, the general solution is $x = n \pi + (-1)^n (-\dfrac{\pi}{6})$ or $x = n \pi + (-1)^n \dfrac{7\pi}{6}$.

Comparing these results with the given options, the solution set is $x = n \pi + (-1)^n \dfrac{7\pi}{6}$ or $x = (2n \pm 1) \dfrac{\pi}{2}$.

Answer: $n \pi+(-1)^n \dfrac{7 \pi}{6}$ or $(2 n \pm 1) \dfrac{\pi}{2}$