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?


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.


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