cos2x = x?

Question

This is quite embarassing... I do the highest level of maths available in high school yet..

basic 3-unit trigonometry eludes me at the moment.

Remind me again how to solve cos2x = x for x?
And yes, I can do it, but I need to know the proper method to be able to solve harder ones coming up in the test.

I suppose this shows that no matter how easy the maths is, if you can't remember it.. it's very challenging..

(General solution is needed.)

I am SO sorry. It's late.

Its cos(2x) = cos(x).

I DO realise how much of a difference this makes. THANKYOU so much for your immediate and elaborate responses. You are so helpful.

AMMENDMENT TO QUESTION ABOVE ^^ AMMENDMENT TO QUESTION ABOVE ^^ AMMENDMENT TO QUESTION ABOVE ^^ AMMENDMENT TO QUESTION ABOVE ^^ AMMENDMENT TO QUESTION ABOVE ^^ AMMENDMENT TO QUESTION ABOVE ^^

Thanks, Geezah. But I'm not sure which to use. Change the cos2x = cos^2(x)-sin^2(x)?
I still can't figure it out from there. The examples seem to lead me to think that I need to use the null factor theorem.

Falzoon, your answer is in radians. You should indicate that =p

A thousand thanks, Geezah. I can rest at night knowing that I can solve this.

I was close, but I recalled nothing about quadratics for trig. Thanks again!

Geezah, you typed = instead of -. For all those people reading it ;_;. You probably have corrected it by now.

but sin^2(x) = 1 - cos^2(x)
and cos^2(x) = 1 - sin^2(x).

Answer 1

cos 2x = cos x
2cos²x - 1 = cos x
2cos²x - cos x - 1 = 0
(2cos x + 1 )(cos x - 1) = 0
cos x = - 1/2 , cos x = 1
x = 120° , 240° , 0° , 360°

Answer 2

Is that cos(2x), or cos^2 (x)? Regardless, there is no real way to analytically solve equations like this. The best you can do is use an approximation method like Newton's Method.

EDIT:
Thanks for the correction, that makes a big difference. You'll want to use some trig identities for substitution. In this case you could use cos(2x) = cos^2(x) - sin^2(x) and sin^2(x) = cos^2(x) = 1 to get
2cos^2(x) - 1 = cos(x)

This can be rewritten as 2cos^2(x) - cos(x) - 1 = 0. Letting p = cos(x), this is 2p^2 - p -1 = 0. Solve this for p using some factoring, then use x = cos^-1(p) to find x.

Answer 3

There must be a typo on the question. The only way to solve these type of questions is by numerical methods, which are not taught in school. You have nothing to worry about- I seriously doubt such a question will be asked in a test or exam.

If you're interested, this website gives one such method for solving such equations:
http://www.ugrad.math.ubc.ca/coursedoc/math100/notes/approx/newton.html
As you can see, it is a first year maths problem at university. Try googling "Newton's Method" or "Numerical Methods" or "Direct iterative method"

Have fun, enjoy, and don't stress!

Answer 4

cos(2x) = cos x
cos^2x -(1-cos^2x)=cos x
so if you call cos x =z
2z^2 -z-1= 0
z=((1+-sqrt(9))/4
z= 1 and z= -1/2
cos x= 1 x= 2k pi
cos x= -1/2 x= 2pi/3 +2k pi and x= 4pi/3 +2k pi
The equation cos2x = x only can be solved by calculus methods
or calculators x=0.514933264662

Answer 5

Newton's method of iteration on f(x) = cos(2x) - x gives :
x = 0.514933264661129

O.K., if it's f(x) = cos^2(x) - x, then :
x = 0.641714370872882

Answer 6

cos(2x) = cos(x)
cos(2x) -cos(x)=0
Use Double angle identity
cos(2x)=cos^2 (x)-sin^2 (x)
cos(2x)=cos^2 (x)-(1-cos^2 (x))
[cos(2x)=2cos^2 (x)-1]
the First equation becomes like
2cos^2 (x)-1-cos(x)=0
rearrange it
2cos^2 (x)-cos(x)-1=0
2cos^2 (x)-2cos(x)+cos(x)-1=0
2cosx(cos(x)-1)+1(cos(x)-1)=0
(2cos(x)+1)(cos(x)-1)=0
x=arccos(-1/2)...........x=arccos(1)
x=0`,60`,300`,360` (Degrees)

Answer 7

I don't remember, there's a property for this ..i think its ..cosxsinx or something along those lines. haha good luck.