Answer
The solution of the equation ${{\cos }^{2}}x+\sin x+1=0$ is $\frac{3\pi }{2}$.
Work Step by Step
Consider the provided equation,
${{\cos }^{2}}x+\sin x+1=0$
Write ${{\cos }^{2}}x=1-{{\sin }^{2}}x$ using the identity ${{\cos }^{2}}x+{{\sin }^{2}}x=1$.
$\begin{align}
& {{\cos }^{2}}x+\sin x+1=0 \\
& 1-{{\sin }^{2}}x+\sin x+1=0 \\
& -{{\sin }^{2}}x+\sin x+2=0 \\
& {{\sin }^{2}}x-\sin x-2=0
\end{align}$
Simplify it further as,
$\begin{align}
& {{\sin }^{2}}x-\sin x-2=0 \\
& {{\sin }^{2}}x-2\sin x+\sin x-2=0 \\
& \left( \sin x-2 \right)\left( \sin x+1 \right)=0
\end{align}$
Equate the factors to zero to obtain the roots.
$\sin x-2\ne 0$.
The maximum value of $\sin x$ is 1.
$\begin{align}
& \left( \sin x-2 \right)\left( \sin x+1 \right)=0 \\
& \sin x+1=0 \\
& \sin x=-1 \\
& x=\frac{3\pi }{2}
\end{align}$
Thus, the solution of the equation ${{\cos }^{2}}x+\sin x+1=0$ is $x=\frac{3\pi }{2}$.
