Question

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

Answer 1

(d) $\left(5\pi /4,7\pi /4\right)$
Given equation:
$$\begin{gathered} {\cos }^2\theta +\sin \theta +1=0 \\ \Rightarrow (1-{\sin }^2\theta )+\sin \theta +1=0 \\ \Rightarrow 2-{\sin }^2\theta +\sin \theta =0 \\ \Rightarrow {\sin }^2\theta -\sin \theta -2=0 \\ \Rightarrow {\sin }^2\theta -2\sin \theta +\sin \theta -2=0 \\ \Rightarrow \sin \theta (\sin \theta -2)+1(\sin \theta -2)=0 \\ \Rightarrow (\sin \theta -2)(\sin \theta +1)=0 \end{gathered}$$
$\Rightarrow \sin \theta -2=0$ or $\sin \theta +1=0$
$\Rightarrow \sin \theta =2$or $\sin \theta =-1$
Now, $\sin \theta =2$ is not possible.
And,

$$\begin{gathered} \sin \theta =-1 \\ \Rightarrow \sin \theta =\sin \frac{3\mathrm{\pi }}{2} \\ \Rightarrow \theta =n\mathrm{\pi }+{\left(-1\right)}^n\frac{3\mathrm{\pi }}{2} \end{gathered}$$

For n = 0, $\theta =\frac{3\mathrm{\pi }}{2}$, for n = 1, $\theta =\frac{7\mathrm{\pi }}{2}$ and so on.
Hence, $\frac{3\mathrm{\pi }}{2}$ lies in the interval $\left(\frac{5\mathrm{\pi }}{4},\frac{7\mathrm{\pi }}{4}\right)$.