# Equation solving

1. Dec 19, 2004

### Eppur si muove

I came upon an equation similar to 0 = (cos x)^2 - x^2 or even 0 = cos x - x and i don't have a clue how to solve it analytically. I tried taking inverse cosine function on both sides but that still doesn't isolate the x. How would you do it?

2. Dec 19, 2004

### dextercioby

The first eq.includes the second one as it can easily be factorized:
$$\cos^{2}x-x^{2}=(\cos x+x)(\cos x-x)=0$$

The two equations obtained as called by the mathematicians "transcendental equations".They are exactly analytically solvable (sorry fo pushing English grammar :tongue2: ) very rarely,and the 2 written above NEVER.
It this case an approximative method is useful and the first one that comes to my mind is the graphic one.Simply plot $\cos x$,$x$ & $-x$ and the solutions to your problem will be the intersection points.

Daniel.

3. Dec 19, 2004

### Eppur si muove

I've approached my original problem differently and now have an equation of the type Acos x = Bcos((pi)x/C)). I have a problem expanding it however. Is there an identity that expands cos(ax) into cos x + f(x) or (cos x)f(x) ?

4. Dec 19, 2004

### ComputerGeek

you could use a taylor series to get a good approximation of cos x and then you should be able to solve for x from that.

that is probably more work than you need to do however, but it is a good place to go if you are desperate.

5. Dec 20, 2004

### Eppur si muove

so there is no ther way? Is it not possible to find an EXACT solution?

6. Dec 20, 2004

### Muzza

See the third sentence of the first reply.