Question

The solution of equations ${\cos }^2\left(\theta \right)+\sin \left(\theta \right)+1=0$ lies in the interval

• A. $\left(-\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{4}\right)$
• B. $\left(\frac{\mathrm{\pi }}{4},\frac{3\mathrm{\pi }}{4}\right)$
• C. $\left(\frac{3\mathrm{\pi }}{4},\frac{5\mathrm{\pi }}{4}\right)$
• D. $\left(\frac{5\mathrm{\pi }}{4},\frac{7\mathrm{\pi }}{4}\right)$

Answer 1

The correct option is D

$\left(\frac{5\mathrm{\pi }}{4},\frac{7\mathrm{\pi }}{4}\right)$

Explanation for the correct option

Given: ${\cos }^2\left(\theta \right)+\sin \left(\theta \right)+1=0$

$$\begin{gathered} \Rightarrow 1-{\sin }^2\left(\theta \right)+\sin \left(\theta \right)+1=0\left[{\sin }^2\theta +{\cos }^2\theta =1\right] \\ \Rightarrow {\sin }^2\left(\theta \right)-\sin \left(\theta \right)-2=0 \\ \Rightarrow {\sin }^2\left(\theta \right)-2\sin \left(\theta \right)+\sin \left(\theta \right)-2=0 \\ \Rightarrow \left(\sin \left(\theta \right)\right)\left[\sin \left(\theta \right)-2\right]+1\left[\sin \left(\theta \right)-2\right]=0 \\ \Rightarrow \left(\sin \left(\theta \right)+1\right)\left(\sin \left(\theta \right)-2\right)=0 \end{gathered}$$

$\sin \left(\theta \right)=2$ is not possible because the range of $\sin \left(\theta \right)$ has a range of $[-1,1]$.

therefore, $\sin \left(\theta \right)+1=0$

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

comparing both sides

$\theta =\frac{3\mathrm{\pi }}{2}\in \left(\frac{5\mathrm{\pi }}{4},\frac{7\mathrm{\pi }}{4}\right)$

Hence, option (D) is correct.