Question

Solve the equation
2 cos^2(x) − cos(x) − 1 = 0, 0 ≤ x ≤ 2π.

Answer 1

This is a quadratic equation in cos(x).
Substitute y = cos(x), and y^2 = cos^2(x) and solve the quadratic equation for y.
Then substitute back cos(x) for y, and solve the trig equation.

Answer 2

okay, that gives me -1/2, and 1. Now to plug that back into the equation just put those values in for x?

Answer 3

No. You don't put them in as x. You originally set y = cos(x) so now that you have two y values, solve for what x's does -1/2 = cos(x). Also look at for what x's does 1 = cos(x). Then look at the interval they gave you.

Answer 4

0, 2pi/3, 4pi/3, and 2pi?

Answer 5

Yes!

Answer 6

awesome, thanks for the help!

Answer 7

\(2 \cos^2x − \cos x − 1 = 0 \)

Let \(y = \cos x\)

\( 2y^2 - y - 1 = 0\)

\( (2y + 1)(y - 1) = 0\)

\( 2y + 1 = 0\) or \(y - 1 = 0\)

\(2y = -1\) or \(y = 1\)

\(y = -\dfrac{1}{2} \) or \(y = 1\)

Now we substitute back.

\(\color{red}{cos x = -\dfrac{1}{2}} \) or \(\color{green}{\cos x = 1}\)

\(\color{red}{x = \dfrac{2\pi}{3}} \) or \(\color{red}{x = \dfrac{4\pi}{3}} \) or \(\color{green}{x = 0} \) or \( \color{green}{x = 2\pi } \)

I used red and green to show which solutions come from which equations.