Question

A solution of the equation ${\cos }^{2}x+\sin x+1=0$, 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) $\left(5\pi /4,7\pi /4\right)$​

Answer 1

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

Given:
$$\begin{gathered} {\cos }^{2}x+\sin x+1=0 \\ \Rightarrow (1-{\sin }^{2}x)+\sin x+1=0 \\ \Rightarrow 1-{\sin }^{2}x+\sin x+1=0 \\ \Rightarrow {\sin }^{2}x-\sin x-2=0 \\ \Rightarrow {\sin }^{2}x-2\sin x+\sin x-2=0 \\ \Rightarrow \sin x(\sin x-2)+1(\sin x-2)=0 \\ \Rightarrow (\sin x-2)(\sin x+1)=0 \end{gathered}$$
$\Rightarrow \sin x-2=0$ or $\sin x+1=0$
$\Rightarrow \sin x=2$ or $\sin x=-1$
$\sin x=2$ is not possible.
$\Rightarrow \sin x=-1$
∴ $\sin x=\sin \frac{3\mathrm{\pi }}{2}$
$\Rightarrow x=n\mathrm{\pi }+(-1{)}^n\frac{3\mathrm{\pi }}{2},\mathrm{n}\in \mathrm{Z}$
The values of $x$ lies in the third and fourth quadrants.
Hence, $x$ lies in $\left(\frac{5\mathrm{\pi }}{4},\frac{7\mathrm{\pi }}{4}\right)$.