A solution of the equation ${\cos }^2\mathrm{\theta }+\sin \mathrm{\theta }+1=0$, lies in the interval
(a) $\left(-\pi /4,\pi /4\right)$
(b) $\left(\pi /4,3\pi /4\right)$
(c) $\left(3\pi /4,5\pi /4\right)$
(d) $\left(5\pi /4,7\pi /4\right)$

(d) $\left(5\pi /4,7\pi /4\right)$

Given:
$$\begin{gathered} {\cos }^2\theta +\sin \theta +1=0 \\ \Rightarrow (1-{\sin }^2\theta )+\sin \theta +1=0 \\ \Rightarrow 1-{\sin }^2\theta +\sin \theta +1=0 \\ \Rightarrow {\sin }^2\theta -\sin \theta -2=0 \\ \Rightarrow {\sin }^2\theta -2\sin \theta +\sin \theta -2=0 \\ \Rightarrow \sin \theta (\sin \theta -2)+1(\sin \theta -2)=0 \\ \Rightarrow (\sin \theta -2)(\sin \theta +1)=0 \end{gathered}$$
$\Rightarrow \sin \theta -2=0$ or $\sin \theta +1=0$
$\Rightarrow \sin \theta =2$ or $\sin \theta =-1$
$\sin \theta =2$ is not possible.
$\Rightarrow \sin \theta =-1$
∴ $\sin \theta =\sin \frac{3\mathrm{\pi }}{2}$
$\Rightarrow \theta =n\mathrm{\pi }+(-1{)}^n\frac{3\mathrm{\pi }}{2},\mathrm{n}\in \mathrm{Z}$

The values of $\theta$ lies in the third and fourth quadrants.
Hence, $\theta$ lies in $\left(\frac{5\mathrm{\pi }}{4},\frac{7\mathrm{\pi }}{4}\right)$.