A solution to two equations ya and yb, in one unknown, x? It generally won't exist. Let me look at them.
ya = (sin((2./(x+2))-3)).^3;
I assume by solution, you mean a point where ya(x) == 0. You are using fzero, after all.
So, what do you know about a cube? z^3 is only zero, when zitself is zero. So we can freely look instead at the cube root. So ya is zero, whenever this expression is zero:
sin((2./(x+2))-3) == 0
When is sin(u) zero? u must be of the form u=n*pi, for integer n. Therefore, we have
2/(x+2) - 3 = n*pi
Solving for x, we find
x = 2/(3+n*pi) - 2
The principal solution lies at n=0, when we have x=-4/3, but there are infinitely many such solutions.
Now, consider yb.
yb = 10*x^2*exp(-2x)*[sin(3*x-pi)+cos(x^2)]
Clearly, x==0 is a solution, so we are done. You only needed one solution anyway. But a product of terms is zero when any member of the product is zero, and not otherwise. exp(-2*x) is never zero, and the last imte I checked 10 is generally never zero either, unless there is some new math I'm unaware of. That leaves the final term.
sin(3*x-pi)+cos(x^2)
I'm feeling too lazy to think for now, so lets throw it at solve and see what falls out.
syms x
solve(sin(3*x-pi)+cos(x^2))
ans =
3/2 - (9 - 2*pi)^(1/2)/2
(9 - 2*pi)^(1/2)/2 + 3/2
- (2*pi + 9)^(1/2)/2 - 3/2
(2*pi + 9)^(1/2)/2 - 3/2
Those appear not to be the only solutions though.
fplot(sin(3*x-pi)+cos(x^2))
refline([0,0])
But the one thing it appears you will probably not find is a solution the two functions have in common.
