Question

Consider the equation:
2cos^2(theta) - 2cos(theta) -1=0.

I've used the quadratic formula to find the degree solutions but it appears that I have done it wrong. I'm trying to find the degree solutions. I know the answer is __?__degrees + 360 degree K and there are two answers but I am having a hard time arriving at the answer. Can someone please help me? I'd appreciate it.

Answer 1

use substitution let

\[u = \cos(\theta)\]


then 2u^2 - 2u - 1 = 0

using the general quadratic formula

\[u=\frac{2\pm \sqrt{(-2)^2 - 4 \times2\times(-1)}}{2\times2}\]

\[u = \frac{2\pm2\sqrt{3}}{4}\]

\[u = \frac{1\pm \sqrt{3}}{2}\]

resubstituting

\[cos(\theta) = \frac{1\pm \sqrt{3}}{2}\]

Answer 2

I get the decimal answers for 1+- Sq.rt3/2 and plug it into the original equation?

Answer 3

Hi jennifer473!

campbell_st has done it for you. still confusion?

Answer 4

I understand the quadratic formula. from the answer I get using the quadratic formula (which is a decimal), do I plug it into the original formula? (2cos^2(theta) - 2cos(theta)-1=0)

Answer 5

no. In question you are asking for solution. so campbell_st done it for you. If you plug your solution in the question then it will be your verification.

Answer 6

So then how do I use the answer formed when using the quadratic equation to get the degrees?

Answer 7

still need help?

Answer 8

Yes please :)

Answer 9

from what I'm gathering you want the angle theta from campbell_st 's solution...
notice there are two solutions...

for the first one, cos(theta) = (1+sqrt3)/2. this does not have a solution because that right side is bigger than 1. so we can throw this one away.

Answer 10

for the other one, cos(theta) = (1-sqrt3)/2, the right side is between -1 and 1 so this has a solution. to find theta,

just take the cosine inverse of that right side: