Question

The 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

Given equation:
$$\begin{gathered} {\cos }^{2}x+\sin x+1=0 \\ \Rightarrow (1-{\sin }^{2}x)+\sin x+1=0 \\ \Rightarrow 2-{\sin }^{2}x+\sin x=0 \\ \Rightarrow {\sin }^{2}x-\sin x-2=0 \\ \Rightarrow {\sin }^{2}x-2\sin x+\sin x-2=0 \\ \Rightarrow \sin x(\sin x-2)+1(\sin x-2)=0 \\ \Rightarrow (\sin x-2)(\sin x+1)=0 \end{gathered}$$
$\Rightarrow \sin x-2=0$ or $\sin x+1=0$
$\Rightarrow \sin x=2$or $\sin x=-1$
Now, $\sin x=2$ is not possible.
And,
$$\begin{gathered} \sin x=-1 \\ \Rightarrow \sin x=\sin \frac{3\mathrm{\pi }}{2} \\ \Rightarrow x=n\mathrm{\pi }+{\left(-1\right)}^n\frac{3\mathrm{\pi }}{2} \end{gathered}$$
For n = 0, $x=\frac{3\mathrm{\pi }}{2}$, for n = 1, ​$x=\frac{7\mathrm{\pi }}{2}$ and so on.
Hence, $\frac{3\mathrm{\pi }}{2}$ lies in the interval $\left(\frac{5\mathrm{\pi }}{4},\frac{7\mathrm{\pi }}{4}\right)$.