The smallest positive angle which satisfies the equation

(a) $\frac{5 \pi}{6}$

Given;

$2 \sin ^{2} x+\sqrt{3} \cos x+1=0$

$\Rightarrow 2\left(1-\cos ^{2} x\right)+\sqrt{3} \cos x+1=0$

$\Rightarrow 2-2 \cos ^{2} x+\sqrt{3} \cos x+1=0$

$\Rightarrow 2 \cos ^{2} x-\sqrt{3} \cos x-3=0$

$\Rightarrow 2 \cos ^{2} x-2 \sqrt{3} \cos x+\sqrt{3} \cos x-3=0$

$\Rightarrow 2 \cos x(\cos x-\sqrt{3})+\sqrt{3}(\cos x-\sqrt{3})=0$

$\Rightarrow(2 \cos x+\sqrt{3})(\cos x-\sqrt{3})=0$

$\Rightarrow 2 \cos x+\sqrt{3}=0$ or, $\cos x-\sqrt{3}=0$

$\therefore \cos x=-\frac{\sqrt{3}}{2} \quad$ or, $\cos x=\sqrt{3}$ is not possible.

$\Rightarrow \cos x=\cos \left(\frac{5 \pi}{6}\right)$

$\Rightarrow x=2 n \pi \pm \frac{5 \pi}{6}, n \in Z$

For $n=0$, the value of $x$ is $\pm \frac{5 \pi}{6}$.

Hence, the smallest positive angle is $\frac{5 \pi}{6}$.