Re: Help with Solve

*To*: mathgroup at smc.vnet.net*Subject*: [mg112337] Re: Help with Solve*From*: Bob Hanlon <hanlonr at cox.net>*Date*: Fri, 10 Sep 2010 04:47:24 -0400 (EDT)

Use Reduce eq1 = n*Sin[x] == (m*v^2)/r; eq2 = n*Cos[x] == m*g; cons = {m > 0, g > 0, r > 0, v > 0}; sys = Join[{eq1, eq2}, cons]; FullSimplify[Reduce[sys, x, Reals], cons] // ToRadicals Element[C[1], Integers] && v == -((n^2*r^2 - g^2*m^2*r^2)^(1/4)/Sqrt[m]) && ((x == 2*(Pi*C[1] + ArcTan[ Sqrt[1 - (2*g*m)/(g*m + n)]]) && n > g*m) || (2*ArcTan[Sqrt[1 - (2*g*m)/(g*m + n)]] + x == 2*Pi*C[1] && g*m + n < 0)) Bob Hanlon ---- Eduardo Cavazos <wayo.cavazos at gmail.com> wrote: ============= Hello! A newb question I'm sure... :-) Here's a couple of equations: eq1 = n*Sin[x] == (m*v^2)/r; eq2 = n*Cos[x] == m*g; The goal is to solve for 'x'. I can do this in a roundabout way via: Solve[eq1 /. Solve[eq2, n], x] I.e. solve eq2 for 'n', substitute this into eq1, and solve the result for 'x'. But this approach seems too "manual". Is there a more straightforward way to carry out the problem? I tried this: Solve[{eq1, eq2}, x] but it doesn't seem to work. What's a good way to go about this? Thanks! Ed