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

Question

asked Mar 4, 2022 229k views

1 Answer

Reachify · Answer 1 · 2022-03-10T04:50:55+0000

Answer:

$\displaystyle x=\left\{0, (2\pi)/(3), (4\pi)/(3)\right\}$

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:

$\displaystyle u=-(1)/(2)\text{ or } u=1$

Back-substitute:

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

Using the unit circle:

$\displaystyle x=\left\{0, (2\pi)/(3), (4\pi)/(3)\right\}$