આપેલ સમીકરણના વ્યાપક ઉકેલ શોધો : $\sin 2 x+\cos x=0$

$\sin 2 x+\cos x=0$

$\Rightarrow 2 \sin x \cos x+\cos x=0$

$\Rightarrow \cos x(2 \sin x+1)=0$

$\Rightarrow \cos x=0 \quad$ or

$2 \sin x+1=0$

Now, $\cos x=0 \Rightarrow \cos x=(2 n+1) \frac{\pi}{2},$ where $n \in Z$

$2 \sin x+1=0$

$\Rightarrow \sin x=\frac{-1}{2}=-\sin \frac{\pi}{6}=\sin \left(\pi+\frac{\pi}{6}\right)=\sin \left(\pi+\frac{\pi}{6}\right)=\sin \frac{7 \pi}{6}$

$\Rightarrow x=n \pi+(-1)^{n} \frac{7 \pi}{6},$ where $n \in Z$

Therefore, the general solution is $(2 n+1) \frac{\pi}{2}$ or $n \pi+(-1)^{n} \frac{7 \pi}{6}, n \in Z$