Help ! two nonlinear equations

cos(3x)-cos(3y)+1 =0
cos(5x)-cos(5y)+1 =0

please, i can not answer these two nonlinear equations by matlab .

Answer 1

[John D'Errico]

Actually, solve has no problems, returning only a set of 6 solutions that all lie in the interval [-pi,pi]. Of course they will not be represented in the form of radicals since those equations are equivalent to a high order (10th degree) polynomial, so vpa is then needed to recover the actual values.

syms x y
xy = solve(cos(3*x)-cos(3*y)+1 ==0,cos(5*x)-cos(5*y)+1 ==0)
xy = 
    x: [6x1 sym]
    y: [6x1 sym]

vpa(xy.x)
ans =
 -0.92557695405354896912598298502098
   3.1415926535897932384626433832795
  0.92557695405354896912598298502098
   1.5707963267948966192313216916398
 -0.92557695405354896912598298502098
  0.92557695405354896912598298502098

Personally, I always like a graphical solution here. One can learn a lot from a picture.

[xx,yy] = meshgrid(linspace(-pi,pi,200));
zz1 = cos(3*xx)-cos(3*yy)+1;
zz2 = cos(5*xx)-cos(5*yy)+1;
contour(xx,yy,zz1,[0 0],'color','r')
hold on
contour(xx,yy,zz2,[0 0],'color','b')
grid on
axis equal

All crossings of the two contour colors are a solution to the nonlinear system. As you can see, there is a nice periodicity to those solutions. There will be infinitely many such solutions of course.

[Muhammed Adel]

thanks sir

Answer 2

[Roger Stafford]

This doesn't solve your equations, but you can gain a better insight into their nature by expanding the cosine expressions in terms of cos(x) and cos(y). Let s = cos(x) and t = cos(y). Then your equations can be expressed:

   (4*s^3-3*s)-(4*t^3-3*t)+1 = 0
   (16*s^5-20*s^3+5*s)-(16*t^5-20*t^3+5*t)+1 = 0

There will be only a finite number of solution pairs to these polynomial equations whereas each solution (s,t) will have infinitely many combinations of corresponding x and y values which are also solutions. Changing to s and t avoids this confusing infinitude of solutions. I suspect 'solve' will be unable to obtain a specific solution to the above, so you may have to resort to numerical methods with 'fsolve'.

[Muhammed Adel]

thanks sir

Answer 3

[John BG]

Muhammed

the expression

    cos(3*x)-cos(3*y)+1=0

is same as

    -2*sin((3*(x+y+x-y))/2)*sin((3*(x+y-(x-y)))/2)=-1
     sin(1.5*x)*sin(1.5*y)=.5

the second equation

sin(2.5*x)*sin(2.5*y)=.5

and the solution are the points:

sin(1.5*x)*sin(1.5*y)-sin(2.5*x)*sin(2.5*y)=0

that d'Errico shows in the 2D graph, only the intersections of blue and red circles that are available directly from the contours:

[c1,h1]=contour(xx,yy,zz1,[0 0],'color','r')
[c2,h2]=contour(xx,yy,zz2,[0 0],'color','b')

the points of the circles are

blue circles

ax1=h1.XData
ay1=h1.YData

red circles

ax2=h2.XData
ay2=h2.YData

the solution:

x1=intersect(ax1,ax2)
y1=intersect(ay1,ay2)

only happens when

    1.5*M*(x+y)=N*2.5*(x-y)

for any positive or negative integer M N

Mr d'Errico, do you agree?

John

[e-mail]

note about trigonometry basics:

[John D'Errico]

I don't see that the relationship

1.5*M*(x+y)=N*2.5*(x-y)

trivially follows from your derivation, nor is it at all useful in predicting solutions.

[Muhammed Adel]

thanks sir