Chapter 14.5, Problem 42PPS

To determine

To find :The solution of the given equation ${\sin }^2\theta =2\sin \theta +3$

Answer to Problem 42PPS

  $(4n-1)90°,n\in \mathbb{Z}$

Explanation of Solution

Given information :

We know that, the given equation is ${\sin }^2\theta =2\sin \theta +3$ .

Formula Used:

  $a{x}^2+bx+c=0\Rightarrow x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation :

The given equation is

  $\begin{array}{l}{\text{sin}}^{\text{2}}\theta -2\sin \theta -3=0\\ \sin \theta =\frac{2\pm \sqrt{4+12}}{2}\\ \sin \theta =\frac{2\pm \sqrt{16}}{2}\\ \sin \theta =1\pm 2\\ \sin \theta =-1or\sin \theta =3\\ \sin \theta =3\text{ is not possible as }-1\le \sin \theta \le 1\\ So,\sin \theta =-1\\ \theta =(4n-1)90°,n\in \mathbb{Z}\end{array}$