Chapter 13.5, Problem 30PPS

To determine

To solve : The given value ${\cos }^2\theta =\frac{1}{4};0°\le \theta \le 360°$

Answer to Problem 30PPS

  $\theta =\frac{\pi }{3},\frac{5\pi }{3}or\theta =\frac{2\pi }{3},\frac{4\pi }{3}$

Explanation of Solution

Given information :

We know that, the given equation is ${\cos }^2\theta =\frac{1}{4};0°\le \theta \le 360°.$

Formula Used:

  $a^2-b^2=(a-b)(a+b)$

Calculation :

The given equation is

  $\begin{array}{l}{\text{cos}}^{\text{2}}\theta -\frac{1}{4}=0\\ {\cos }^2\theta -{\left(\frac{1}{2}\right)}^2=0\\ \left(\cos \theta -\frac{1}{2}\right)\left(\cos \theta +\frac{1}{2}\right)=0\\ \cos \theta =\frac{1}{2}or\cos \theta =-\frac{1}{2}\\ \theta =\frac{\pi }{3},\frac{5\pi }{3}or\theta =\frac{2\pi }{3},\frac{4\pi }{3}\end{array}$