Chapter 13, Problem 41SGR

To determine

To find : The solution of $4{\cos }^2\theta -4\cos \theta +1=0;0\le \theta \le 2\pi$

Answer to Problem 41SGR

  $\theta =\frac{\pi }{3}as\frac{5\pi }{3}$

Explanation of Solution

Given information :

We know that, the given equation is $4{\cos }^2\theta -4\cos \theta +1=0.$

Formula Used :

  ${(a-b)}^2=a^2-2ab+b^2$

Calculation :

The given equation is

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