Question

Use any method to solve the equation over the interval [0,2pi). Express solutions to the nearest hundredth cos2x-cosx=0

Answer 1

One of the identities for cos(2x) is completely in terms of cos(x):

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

Substitute that in, and you will have an equation of quadratic type. Solve using methods for quadratic equations.

Does that help?

Answer 2

After using the method for quadratic equations would I end up with x=1/2 & x=0?

Answer 3

I don't think you got x=1/2...... double check that. Remember, when you solve the quadratic, you are solving for "cosx", not for x. Then you have a simple cosine equation that you need to solve for x.

Also, check your signs from solving the quadratic.

Answer 4

Actually, I should say, when you solve the quadratic you get TWO solutions for cosx, so you have 2 simple cosine equations to solve.

That is, solving the quadratic gives you numerical values for cos(x), not solutions for x. To find the solution(s) for x, you have to solve the cosine equations that fall out of the quadratic.

Answer 5

I'm not sure if I understand entirely. Would you mind helping me setp by step? I would like to learn how to solve these problems. If you wouldn't mind showing me a set up of the problem, that would be greatly appreciated!

Answer 6

You have cos2x-cosx=0

And we have an identity, \[\cos(2x)=2\cos^2(x)-1\]

right? So substituting that in, we get:

\[\ 2cos^2(x)-1-cos(x)=0\]

or, rearranging so that it "looks" more quadratic-like, we have:

\[\ 2cos^2(x)-cos(x)-1=0\]

Answer 7

If you prefer, at this point you can make a substitution, \[\ cos(x)=u\]
and write:
\[\ 2u^2-u-1=0\]
this step is not really necessary, but some prefer to look at an equation in u, rather than in cos(x).

Answer 8

Now solve that quadratic. You get 2 solutions for u (which are really solutions for cos(x)).

You should get cos(x)=-1/2 or cos(x)=1

Now, solve EACH of those, using the inverse function (or just by knowing what angle(s) have those cosine values - they are both among the "common angles" so you should know them)

Answer 9

Oh, I see that your problem says "to the nearest hundreth", so you don't really even need to know the exact values - apparently you are allowed to use a calculator? Just remember that the calculator's inverse function will only give one result. For cos(x)=-1/2, there are 2 solutions on the interval given.

Answer 10

I see where my problem was. When assembling my equation, I added a (+1) out of thin air. EX: 2cos^2x-1-cosx+1. I was randomly adding numbers and confusing myself. Thank you.

Answer 11

Yeah, randomly adding numbers would definitely confuse the problem. :) You're welcome.

Answer 12

SOrry Debbie, I came across another problem. Instead of your asnwers cos(x) =- 1/2 or cos(x)=1. I've gotten cos(x)= 5/2 or cos(x)=2

Answer 13

Did you factor?

\[ 2u^2-u-1=0\]\[ (2u+1)(u-1)=0\]
u=-1/2 or u=1

Answer 14

No I did not. I used the quadratic formula.