0
$\begingroup$

I have two solutions to quadratic equations, based on the quadratic formula. The solutions are equivalent. Additionally, one of the variables (Tx, Ty) in both sides of the equation is a function of an angle x. I want to solve for that angle x.

I tried plugging into Mathematica, but it is taking forever to solve and it won't even tell me if a solution exists.

(v+sqrt(v^2+2*d*T*cos(x)))*sin(x) = cos(x)*(w+sqrt(w^2+2*g*T*sin(x)))

Does a solution exist for x?

$\endgroup$
1
$\begingroup$

There are three easy solution: $x=-\tfrac{\pi}{2},$ $x=0$ and $x=\tfrac{\pi}{2}$. The others are given implicitly as

arctan(-w^2/(g*T)+RootOf((g^2+d^2)*_Z^4+(-4*w*d^2+4*g*v*d)*_Z^3+(4*g^2*v^2-2*g^2*w^2+6*w^2*d^2-12*g*v*w*d)*_Z^2+(-4*w^3*d^2-8*g^2*v^2*w+12*g*v*w^2*d)*_Z+4*g^2*v^2*w^2+w^4*d^2-4*g*v*w^3*d+w^4*g^2-4*g^4*T^2)^2/(g*T), (-2*g*v*w+w^2*d)/(g^2*T)+(2*g*v-2*w*d)*RootOf((g^2+d^2)*_Z^4+(-4*w*d^2+4*g*v*d)*_Z^3+(4*g^2*v^2-2*g^2*w^2+6*w^2*d^2-12*g*v*w*d)*_Z^2+(-4*w^3*d^2-8*g^2*v^2*w+12*g*v*w^2*d)*_Z+4*g^2*v^2*w^2+w^4*d^2-4*g*v*w^3*d+w^4*g^2-4*g^4*T^2)/(g^2*T)+d*RootOf((g^2+d^2)*_Z^4+(-4*w*d^2+4*g*v*d)*_Z^3+(4*g^2*v^2-2*g^2*w^2+6*w^2*d^2-12*g*v*w*d)*_Z^2+(-4*w^3*d^2-8*g^2*v^2*w+12*g*v*w^2*d)*_Z+4*g^2*v^2*w^2+w^4*d^2-4*g*v*w^3*d+w^4*g^2-4*g^4*T^2)^2/(g^2*T))

where the "RootOf" uses _Z as the variable. So it seems to think that the other solutions are given by combining the roots to quartic polynomials and then taking the inverse tangent of these.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.