Chapter 14.5, Problem 53PPS

To determine

To find :The solution of $1-{\sin }^2\theta -\cos \theta =\frac{3}{4}$

Answer to Problem 53PPS

  $\theta =2n\pi \pm \frac{2\pi }{3}$

Explanation of Solution

Given information :

We know that, the given equation is $1-{\sin }^2\theta -\cos \theta =\frac{3}{4}$

Formula Used:

  ${\sin }^2\theta =1-{\cos }^2\theta$

Calculation :

The given equation is

  $\begin{array}{l}1-{\sin }^2\theta -\cos \theta =\frac{3}{4}\\ 1-(1-{\cos }^2\theta )-\cos \theta =\frac{3}{4}\\ 4{\cos }^2\theta -4\cos \theta -3=0\\ 4{\cos }^2\theta -6\cos \theta +2\cos \theta -3=0\\ 2\cos \theta (2\cos \theta -3)+1(2\cos \theta -3)=0\\ (2\cos \theta +1)(2\cos \theta -3)=0\\ \cos \theta =-\frac{1}{2}or\cos \theta =\frac{3}{2}\\ \text{As}-1\le \cos \theta \le 1so\cos \theta =\frac{3}{2}\text{ is not possible}\text{.}\\ \cos \theta =\cos \left(\frac{2\pi }{3}\right)\\ \theta =2n\pi \pm \frac{2\pi }{3}\end{array}$