matlab equation !!!! solve('tan(2.6433*x)=(1-x^2-(3.4/3.66)^2)^.5/x')

[W_Heisenberg]

the equation as follows:
tan(2.6433*x)=(1-x^2-(3.4/3.66)^2)^.5 /x

this should not be so hard to solve by matlab, but matlab never gives me anwser, anyone knows why?

my codes :
solve('tan(2.6433*x)=(1-x^2-(3.4/3.66)^2)^.5/x')

 

[albbg]

It could be your variable 'x' isn't in the default variables. Try to specify the variable using solve(expr,var), that is:

solve('tan(2.6433*x)=(1-x^2-(3.4/3.66)^2)^.5/x',x)

or try enabling the symbolic solver:

syms x
solve('tan(2.6433*x)=(1-x^2-(3.4/3.66)^2)^.5/x')

 

[_Eduardo_]

the equation as follows:
tan(2.6433*x)=(1-x^2-(3.4/3.66)^2)^.5 /x

this should not be so hard to solve by matlab, but matlab never gives me anwser, anyone knows why?

my codes :
solve('tan(2.6433*x)=(1-x^2-(3.4/3.66)^2)^.5/x')

Rewrite your equation for some "numerically friendly" (avoiding square root)

tan(2.6433*x)^2 = (1-x^2-(3.4/3.66)^2)/x^2

x = 0.2759735729

 

[W_Heisenberg]

Wow, buddy

Square root should be avoided in matlab?

Is this your suggestion for all the equations? or only this one?

Besides, thanks for your help !!!!

 

[_Eduardo_]

Wow, buddy
Square root should be avoided in matlab?
Is this your suggestion for all the equations? or only this one?

When solving something for numerical methods you have to be careful with the function domain, because if an iteration falls outside (eg. complex values in real function, division by zero or near zero , etc ​​), the software will have to do special handling of that situation.

With sqrt that occurs on negative values​​, as often the treatment given by the software is unknown, the alternative is modify (if possible) the equation.

In the previous message I did not have near a PC with Matlab installed and try with Derive6.
With tan(2.6433*x) = (1-x^2 - (3.4/3.66)^2)^.5/x does not converge
but (tan(2.6433*x))^2 = (1-x ^ 2 - (3.4/3.66)^2)/x^2
or (x*tan(2.6433*x))^2 = 1-x^2 - (3.4 / 3.66)^2 works fine


Minutes ago I did with Matlab and solve always fail (I don't know why) :sad:
With fsolve(funct, x0) converges in all ways, but only with initial values ​​(x0) relatively close to the root.

 

[W_Heisenberg]

I actually failed thousands of times. It keeps saying left of equal sign cant be assigned.

Man, your explaination makes sense. I do believe this is a problem of the codes.

I mean, it could be the software engineer defined the domain in a certain range. In order to use the function properly, we should read the mannual carefully.

In fact, originally my question is a function of x, rather this x = cos(z), and solving for z.

It is worse if I do not use x to represent cos(z). Nonlinear equations codes could be even more complex.

Yet now, I really thank you, buddy. With your help, I finally could solve something.