How to solve these 3 equations with 3 variables which are in radian.

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The variables to find are t1,t2,t3 which are supposed to be in radian
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John D'Errico
John D'Errico el 29 de Abr. de 2023
Depending on how you want to approach the problem, and what tools you have available...
help solve
--- help for sym/solve --- SOLVE Symbolic solution of algebraic equations. S = SOLVE(eqn1,eqn2,...,eqnM,var1,var2,...,varN) S = SOLVE(eqn1,eqn2,...,eqnM,var1,var2,...,varN,'ReturnConditions',true) [S1,...,SN] = SOLVE(eqn1,eqn2,...,eqnM,var1,var2,...,varN) [S1,...,SN,params,conds] = SOLVE(eqn1,...,eqnM,var1,var2,...,varN,'ReturnConditions',true) The eqns are symbolic expressions, equations, or inequalities. The vars are symbolic variables specifying the unknown variables. If the expressions are not equations or inequalities, SOLVE seeks zeros of the expressions. Otherwise SOLVE seeks solutions. If not specified, the unknowns in the system are determined by SYMVAR, such that their number equals the number of equations. If no analytical solution is found, a numeric solution is attempted; in this case, a warning is printed. Three different types of output are possible. For one variable and one output, the resulting solution is returned, with multiple solutions to a nonlinear equation in a symbolic vector. For several variables and several outputs, the results are sorted in the same order as the variables var1,var2,...,varN in the call to SOLVE. In case no variables are given in the call to SOLVE, the results are sorted in lexicographic order and assigned to the outputs. For several variables and a single output, a structure containing the solutions is returned. SOLVE(...,'ReturnConditions', VAL) controls whether SOLVE should in addition return a vector of all newly generated parameters to express infinite solution sets and about conditions on the input parameters under which the solutions are correct. If VAL is TRUE, parameters and conditions are assigned to the last two outputs. Thus, if you provide several outputs, their number must equal the number of specified variables plus two. If you provide a single output, a structure is returned that contains two additional fields 'parameters' and 'conditions'. No numeric solution is attempted even if no analytical solution is found. If VAL is FALSE, then SOLVE may warn about newly generated parameters or replace them automatically by admissible values. It may also fall back to the numerical solver. The default is FALSE. SOLVE(...,'IgnoreAnalyticConstraints',VAL) controls the level of mathematical rigor to use on the analytical constraints of the solution (branch cuts, division by zero, etc). The options for VAL are TRUE or FALSE. Specify FALSE to use the highest level of mathematical rigor in finding any solutions. The default is FALSE. SOLVE(...,'PrincipalValue',VAL) controls whether SOLVE should return multiple solutions (if VAL is FALSE), or just a single solution (when VAL is TRUE). The default is FALSE. SOLVE(...,'IgnoreProperties',VAL) controls if SOLVE should take assumptions on variables into account. VAL can be TRUE or FALSE. The default is FALSE (i.e., take assumptions into account). SOLVE(...,'Real',VAL) allows to put the solver into "real mode." In "real mode," only real solutions such that all intermediate values of the input expression are real are searched. VAL can be TRUE or FALSE. The default is FALSE. SOLVE(...,'MaxDegree',n) controls the maximum degree of polynomials for which explicit formulas will be used during the computation. n must be a positive integer. The default is 3. Example 1: syms p x r solve(p*sin(x) == r) chooses 'x' as the unknown and returns ans = asin(r/p) pi - asin(r/p) Example 2: syms x y [Sx,Sy] = solve(x^2 + x*y + y == 3,x^2 - 4*x + 3 == 0) returns Sx = 1 3 Sy = 1 -3/2 Example 3: syms x y S = solve(x^2*y^2 - 2*x - 1 == 0,x^2 - y^2 - 1 == 0) returns the solutions in a structure. S = x: [8x1 sym] y: [8x1 sym] Example 4: syms a u v [Su,Sv] = solve(a*u^2 + v^2 == 0,u - v == 1) regards 'a' as a parameter and solves the two equations for u and v. Example 5: syms a u v w S = solve(a*u^2 + v^2,u - v == 1,a,u) regards 'v' as a parameter, solves the two equations, and returns S.a and S.u. When assigning the result to several outputs, the order in which the result is returned depends on the order in which the variables are given in the call to solve: [U,V] = solve(u + v,u - v == 1, u, v) assigns the value for u to U and the value for v to V. In contrast to that [U,V] = solve(u + v,u - v == 1, v, u) assigns the value for v to U and the value of u to V. Example 6: syms a u v [Sa,Su,Sv] = solve(a*u^2 + v^2,u - v == 1,a^2 - 5*a + 6) solves the three equations for a, u and v. Example 7: syms x S = solve(x^(5/2) == 8^(sym(10/3))) returns all three complex solutions: S = 16 - 4*5^(1/2) - 4 + 4*2^(1/2)*(5 - 5^(1/2))^(1/2)*i - 4*5^(1/2) - 4 - 4*2^(1/2)*(5 - 5^(1/2))^(1/2)*i Example 8: syms x S = solve(x^(5/2) == 8^(sym(10/3)), 'PrincipalValue', true) selects one of these: S = - 4*5^(1/2) - 4 + 4*2^(1/2)*(5 - 5^(1/2))^(1/2)*i Example 9: syms x S = solve(x^(5/2) == 8^(sym(10/3)), 'IgnoreAnalyticConstraints', true) ignores branch cuts during internal simplifications and, in this case, also returns only one solution: S = 16 Example 10: syms x S = solve(sin(x) == 0) returns 0 S = solve(sin(x) == 0, 'ReturnConditions', true) returns a structure expressing the full solution: S.x = k*pi S.parameters = k S.conditions = in(k, 'integer') Example 11: syms x y real [S, params, conditions] = solve(x^(1/2) = y, x, 'ReturnConditions', true) assigns solution, parameters and conditions to the outputs. In this example, no new parameters are needed to express the solution: S = y^2 params = Empty sym: 1-by-0 conditions = 0 <= y Example 12: syms a x y [x0, y0, params, conditions] = solve(x^2+y, x, y, 'ReturnConditions', true) generates a new parameter z to express the infinitely many solutions. This z can be any complex number, both solutions are valid without restricting conditions: x0 = -(-z)^(1/2) (-z)^(1/2) y0 = z z params = z conditions = true true Example 13: syms t positive solve(t^2-1) ans = 1 solve(t^2-1, 'IgnoreProperties', true) ans = 1 -1 Example 14: solve(x^3-1) returns all three complex roots: ans = 1 - 1/2 + (3^(1/2)*i)/2 - 1/2 - (3^(1/2)*i)/2 solve(x^3-1, 'Real', true) only returns the real root: ans = 1 See also DSOLVE, SUBS. Documentation for sym/solve doc sym/solve Other uses of solve femodel/solve FunctionApproximation.Problem/solve optim.problemdef.OptimizationProblem/solve pde.femodel/solve simscape.multibody.KinematicsSolver/solve
help fsolve
FSOLVE solves systems of nonlinear equations of several variables. FSOLVE attempts to solve equations of the form: F(X) = 0 where F and X may be vectors or matrices. FSOLVE implements three different algorithms: trust region dogleg, trust region, and Levenberg-Marquardt. Choose one via the option Algorithm: for instance, to choose trust region, set OPTIONS = optimoptions('fsolve','Algorithm','trust-region'), and then pass OPTIONS to FSOLVE. X = FSOLVE(FUN,X0) starts at the matrix X0 and tries to solve the equations in FUN. FUN accepts input X and returns a vector (matrix) of equation values F evaluated at X. X = FSOLVE(FUN,X0,OPTIONS) solves the equations with the default optimization parameters replaced by values in OPTIONS, an argument created with the OPTIMOPTIONS function. See OPTIMOPTIONS for details. Use the SpecifyObjectiveGradient option to specify that FUN also returns a second output argument J that is the Jacobian matrix at the point X. If FUN returns a vector F of m components when X has length n, then J is an m-by-n matrix where J(i,j) is the partial derivative of F(i) with respect to x(j). (Note that the Jacobian J is the transpose of the gradient of F.) X = FSOLVE(PROBLEM) solves system defined in PROBLEM. PROBLEM is a structure with the function FUN in PROBLEM.objective, the start point in PROBLEM.x0, the options structure in PROBLEM.options, and solver name 'fsolve' in PROBLEM.solver. Use this syntax to solve at the command line a problem exported from OPTIMTOOL. [X,FVAL] = FSOLVE(FUN,X0,...) returns the value of the equations FUN at X. [X,FVAL,EXITFLAG] = FSOLVE(FUN,X0,...) returns an EXITFLAG that describes the exit condition. Possible values of EXITFLAG and the corresponding exit conditions are listed below. See the documentation for a complete description. 1 FSOLVE converged to a root. 2 Change in X too small. 3 Change in residual norm too small. 4 Computed search direction too small. 0 Too many function evaluations or iterations. -1 Stopped by output/plot function. -2 Converged to a point that is not a root. -3 Trust region radius too small (Trust-region-dogleg). [X,FVAL,EXITFLAG,OUTPUT] = FSOLVE(FUN,X0,...) returns a structure OUTPUT with the number of iterations taken in OUTPUT.iterations, the number of function evaluations in OUTPUT.funcCount, the algorithm used in OUTPUT.algorithm, the number of CG iterations (if used) in OUTPUT.cgiterations, the first-order optimality (if used) in OUTPUT.firstorderopt, and the exit message in OUTPUT.message. [X,FVAL,EXITFLAG,OUTPUT,JACOB] = FSOLVE(FUN,X0,...) returns the Jacobian of FUN at X. Examples FUN can be specified using @: x = fsolve(@myfun,[2 3 4],optimoptions('fsolve','Display','iter')) where myfun is a MATLAB function such as: function F = myfun(x) F = sin(x); FUN can also be an anonymous function: x = fsolve(@(x) sin(3*x),[1 4],optimoptions('fsolve','Display','off')) If FUN is parameterized, you can use anonymous functions to capture the problem-dependent parameters. Suppose you want to solve the system of nonlinear equations given in the function myfun, which is parameterized by its second argument c. Here myfun is a MATLAB file function such as function F = myfun(x,c) F = [ 2*x(1) - x(2) - exp(c*x(1)) -x(1) + 2*x(2) - exp(c*x(2))]; To solve the system of equations for a specific value of c, first assign the value to c. Then create a one-argument anonymous function that captures that value of c and calls myfun with two arguments. Finally, pass this anonymous function to FSOLVE: c = -1; % define parameter first x = fsolve(@(x) myfun(x,c),[-5;-5]) See also OPTIMOPTIONS, LSQNONLIN, @. Documentation for fsolve doc fsolve

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Torsten
Torsten el 29 de Abr. de 2023
Editada: Torsten el 29 de Abr. de 2023
syms t1 t2 t3
eqn1 = 140 + 140*cos(t1)*sin(t2)*sin(t3)-140*cos(t1)*cos(t2)*cos(t3)-180*cos(t1)*sin(t2)==0;
eqn2 = 140*sin(t1)*sin(t2)*sin(t3)-140*sin(t1)*cos(t2)*cos(t3)-180*sin(t1)*sin(t2)==0;
eqn3 = -297 - 140*cos(t2)*sin(t3)-140*sin(t2)*cos(t3)+180*cos(t2)+117==0;
sol = solve([eqn1 eqn2 eqn3],'ReturnConditions',1);
sol.t1
ans = 
sol.t2
ans = 
sol.t3
ans = 
sol.conditions
ans = 
sol.parameters
ans = 
  5 comentarios
Torsten
Torsten el 29 de Abr. de 2023
Editada: Torsten el 29 de Abr. de 2023
If there is no analytical solution as for your first set of equations, all depends on the initial guess for the parameters.
Try
sol = vpasolve([eqn1 eqn2 eqn3],[t1 t2 t3],[0.78 0.78 0.78]);
Ronak Narkhede
Ronak Narkhede el 29 de Abr. de 2023
I'll figure it out after some documentation reading, you have been of immense help thank you sir.

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