Answer:
[tex]\displaystyle x=\left\{0, \frac{2\pi}{3}, \frac{4\pi}{3}\right\}[/tex]
Step-by-step explanation:
We want to solve the equation:
[tex]-2\cos^2(x)=-\cos(x)-1[/tex]
Over the interval [0, 2π).
First, notice that this is in quadratic form. So, to make things simpler, we can let u = cos(x). Substitute:
[tex]-2u^2=-u-1[/tex]
Rearrange:
[tex]2u^2-u-1=0[/tex]
Factor:
[tex](2u+1)(u-1)=0[/tex]
Zero Product Property:
[tex]2u+1=0\text{ or } u-1=0[/tex]
Solve for each case:
[tex]\displaystyle u=-\frac{1}{2}\text{ or } u=1[/tex]
Back-substitute:
[tex]\displaystyle \cos(x)=-\frac{1}{2}\text{ or } \cos(x)=1[/tex]
Using the unit circle:
[tex]\displaystyle x=\left\{0, \frac{2\pi}{3}, \frac{4\pi}{3}\right\}[/tex]
