Question

In Problems 57– 80, solve each equation on the interval $0\le \theta \le 2\mathrm{\pi }$

$2{\cos }^2\theta +\cos \theta -1=0$

Short answer

The solution set of the given equation is$\left\{\mathrm{\pi },\frac{\mathrm{\pi }}{3},\frac{5\mathrm{\pi }}{3}\right\}$.

Step-by-step solution

Step 1. Given Information 

In the given problems we have to solve each equation on the interval $0\le \theta \le 2\mathrm{\pi }$.

$2{\cos }^2\theta +\cos \theta -1=0$

Step 2. This equation is a quadratic equation (in cosθ) that can be factored. 

$$\begin{gathered} 2{\cos }^2\theta +\cos \theta -1=0 \\ 2{\cos }^2\theta +(2-1)\cos \theta -1=0 \\ 2{\cos }^2\theta +2\cos \theta -\cos \theta -1=0 \\ 2\cos \theta (\cos \theta +1)-1(\cos \theta +1)=0 \\ (2\cos \theta -1)(\cos \theta +1)=0 \end{gathered}$$

Step 3. Use the Zero-Product Property. 

$$\begin{gathered} 2\cos \theta -1=0\mathrm{or}\cos \theta +1=0 \\ 2\cos \theta -1+1=0+1\mathrm{or}\cos \theta +1-1=0-1 \\ 2\cos \theta =1\mathrm{or}\cos \theta =-1 \\ \frac{2}{2}\cos \theta =\frac{1}{2}\mathrm{or}\cos \theta =-1 \\ \cos \theta =\frac{1}{2}\mathrm{or}\cos \theta =-1 \end{gathered}$$

Step 4. Solving each equation in the interval [0,2π), we obtain 

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

So the solution set is$\left\{\mathrm{\pi },\frac{\mathrm{\pi }}{3},\frac{5\mathrm{\pi }}{3}\right\}$.