Chapter 14.5, Problem 2ACYP

To determine

To solve :The given value $\cos 2\theta +\cos \theta +1=0$

Answer to Problem 2ACYP

  $\theta =(2n+1)\frac{\pi }{2},2n\pi \pm \frac{2\pi }{3},n\in z$

Explanation of Solution

Given information :

We know that, the given expression is $\cos 2\theta +\cos \theta +1=0.$

Formula Used :

  $1+\cos 2\theta =2{\cos }^2\theta$

Calculation :

The given equation is

  $\begin{array}{l}\Rightarrow \cos 2\theta +\cos \theta +1=0\\ \Rightarrow 1+\cos 2\theta +\cos \theta =0\\ \Rightarrow 2{\cos }^2\theta +\cos \theta =0\\ \Rightarrow \cos \theta (2\cos \theta +1)=0\\ \Rightarrow \cos \theta =0or2\cos \theta +1=0\\ \theta =(2n+1)\frac{\pi }{2},n\in zor2\cos \theta +1=0\\ or\cos \theta =\frac{1}{2}\\ or\cos \theta =-\cos \frac{\pi }{3}\\ or\cos \theta =\cos \frac{2\pi }{3}\\ \theta =2n\pi \pm \frac{2\pi }{3},n\in z\end{array}$