Explanation:

- Given equation: $2{\cos }^2\theta +\sin \theta -2=0$.

- Use the identity ${\cos }^2\theta =1-{\sin }^2\theta$.

- Substitute: $2(1-{\sin }^2\theta )+\sin \theta -2=0$.

- Simplify: $2-2{\sin }^2\theta +\sin \theta -2=0$.

- This simplifies to: $-2{\sin }^2\theta +\sin \theta =0$.

- Factor: $\sin \theta (1-2\sin \theta )=0$.

- Solutions: $\sin \theta =0$ or $1-2\sin \theta =0$.

- $\sin \theta =0$ yields no solution in $0<\theta \le \pi /2$.

- $1-2\sin \theta =0$ gives $\sin \theta =\frac{1}{2}$.

- $\theta =\frac{\pi }{6}$;

- Possible $\theta$ value: $\theta =\frac{\pi }{6}$, $\sin \theta =\frac{1}{2}$.

- Correct Option: Option 1 - $\frac{\pi }{6}$