Question

Solve the equation for exact solutions over the interval​ [0, 2π​).

Answer 1

Answer:

$\displaystyle x=\left\{0, \frac{2\pi}{3}, \frac{4\pi}{3}\right\}$

Step-by-step explanation:

We want to solve the equation:

$-2\cos^2(x)=-\cos(x)-1$

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:

$-2u^2=-u-1$

Rearrange:

$2u^2-u-1=0$

Factor:

$(2u+1)(u-1)=0$

Zero Product Property:

$2u+1=0\text{ or } u-1=0$

Solve for each case:

$\displaystyle u=-\frac{1}{2}\text{ or } u=1$

Back-substitute:

$\displaystyle \cos(x)=-\frac{1}{2}\text{ or } \cos(x)=1$

Using the unit circle:

$\displaystyle x=\left\{0, \frac{2\pi}{3}, \frac{4\pi}{3}\right\}$