Chapter 14.5, Problem 15CYU

To determine

To find :The value of $\theta$ in degree.

Answer to Problem 15CYU

  $\theta =90°and-90°.$

Explanation of Solution

Given information :

We know that, The given information is $\text{cos 2}\theta -{\sin }^2\theta +2=0.$

Calculation :

  $\begin{array}{l}As,\cos 2\theta =1-2{\sin }^2\theta \\ \Rightarrow \cos 2\theta -{\sin }^2\theta +2=0\\ \Rightarrow 1-2{\sin }^2\theta -{\sin }^2\theta +2=0\\ \Rightarrow -3{\sin }^2\theta +3=0\\ \Rightarrow 3{\sin }^2\theta =3\\ \Rightarrow {\sin }^2\theta =1\\ \Rightarrow \sin \theta =\pm 1\\ \Rightarrow \sin \theta =\pm \sin \frac{\pi }{2}\\ \Rightarrow q=90°and-90°\end{array}$