solving equation

[xc630]

Hi I need to solve this following equation:

cos^2x - 2 cosx-1=0

I did these steps:

cos^2x- 2cosx -1 =0
cos^2x - 2cosx +(1) = 1+(1)
(cos x-1) (cosx-1) = 2
cos x=3

I completed the square and got cos x=3 but this answer does not work on the calculator. Did I do something wrong or how do I fix my answer?

 

[Unco]

When in doubt, use a dummy variable.

Let \(\displaystyle \L t = \cos{x}\)

The equation becomes
\(\displaystyle \L t^2 \, - \, 2t \, - \, 1 \, = \, 0\)

We can then solve for t by completing the square (or using the quadratic formula), and substituting back cos(x) for t.

Remember that any solution having cos(x) greater than 1 (or less than -1) can be thrown out.

 

[xc630]

I completed the square and the only answer I got was 3. that is not possible so does that mean the problem has no answer?

 

[Unco]

Let's try it:

\(\displaystyle \L t^2 \, - \, 2t \, - \, 1 \, = \, 0\)

\(\displaystyle \L t^2 \, - \, 2t \, = \, 1\)

\(\displaystyle \L (t - 1)^2 \, - \, 1 \, = \, 1\)

\(\displaystyle \L (t-1)^2 \, = \, 2\)

\(\displaystyle \L t \, = \, 1 \, \pm \, \sqrt{2}\)

 

[Guest]

You got 3? Just seen Unco beat me to it :lol: he's totally right, one is a solution one is not.

 

[xc630]

I dont get how you got

(t-1)^2 -1 = 1 from t^2-2t=1

 

[xc630]

Nevermind

I did it this way

cosx^2 - 2 cos x =1

cos x (cos x-2)=1

cosx=1

cos x=1/2

 

[Guest]

Unfortunately you can't do that unless the RHS =0

 

[xc630]

Ok then revert back to my preious question on how that equation was derived

 

[Guest]

Like Unco said;

\(\displaystyle t^2 -2t -1=0\) Do you see where this comes from?

I always add this extrs step for clarity:

\(\displaystyle (t -1)^2 =t^2 -2t +1\)

So we see that we can -2 from both sides to get the original equation we wanted

\(\displaystyle (t -1)^2 -2 =t^2 -2t -1\)

\(\displaystyle (t -1)^2 -2 =0\)

 

[xc630]

Thank you for the clear up. I got cos x= 1- SQRT2

That leaves an answer of 1.99 in radians