Solve Equation 0 = (cos x)^2 - x^2 or 0 = cos x - x

[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?

 

[dextercioby]

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?

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.

 

[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) ?

 

[ComputerGeek]

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?

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.

 

[Eppur si muove]

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

 

[Muzza]

See the third sentence of the first reply.