# How to solve the over-determined equations?

That is,

I want to solve these equations below:

1

-(Sin[\[Delta]] Cos[\[Lambda]] Sin[2 \[Phi]] +
Sin[2 \[Delta]] Sin[\[Lambda]] Sin[\[Phi]]^2)==0


2

(Sin[\[Delta]] Cos[\[Lambda]] Sin[2 \[Phi]] -
Sin[2 \[Delta]] Sin[\[Lambda]] Cos[\[Phi]]^2)==0


3

-(Sin[\[Delta]] Cos[\[Lambda]] Sin[2 \[Phi]] +
Sin[2 \[Delta]] Sin[\[Lambda]] Sin[\[Phi]]^2)+(Sin[\[Delta]] Cos[\[Lambda]] Sin[2 \[Phi]] -
Sin[2 \[Delta]] Sin[\[Lambda]] Cos[\[Phi]]^2)==0


4

(Sin[\[Delta]] Cos[\[Lambda]] Cos[2 \[Phi]] +
1/2 Sin[2 \[Delta]] Sin[\[Lambda]] Sin[2 \[Phi]])==1


5

-(Cos[\[Delta]] Cos[\[Lambda]] Cos[\[Phi]] +
Cos[2 \[Delta]] Sin[\[Lambda]] Sin[\[Phi]])==0


6

-(Cos[\[Delta]] Cos[\[Lambda]] Sin[\[Phi]] -
Cos[2 \[Delta]] Sin[\[Lambda]] Cos[\[Phi]])==0


constraint condition:

 0 <= \[Phi] < 2 \[Pi] && 0 <= \[Delta] <= \[Pi]/2 &&
0 <= \[Lambda] < 2 \[Pi]


solve

\[Phi], \[Delta], \[Lambda]


Solve[{-(Sin[δ] Cos[λ] Sin[2 ϕ]+Sin[2 δ] Sin[λ] Sin[ϕ]^2)==0,
(Sin[δ] Cos[λ] Sin[2 ϕ]-Sin[2 δ] Sin[λ] Cos[ϕ]^2)==0,
-(Sin[δ] Cos[λ] Sin[2 ϕ]+Sin[2 δ] Sin[λ] Sin[ϕ]^2)+(Sin[δ] Cos[λ] Sin[2 ϕ]-Sin[2 δ] Sin[λ] Cos[ϕ]^2)==0,
(Sin[δ] Cos[λ] Cos[2 ϕ]+1/2 Sin[2 δ] Sin[λ] Sin[2 ϕ])==1,
-(Cos[δ] Cos[λ] Cos[ϕ]+Cos[2 δ] Sin[λ] Sin[ϕ])==0,
-(Cos[δ] Cos[λ] Sin[ϕ]-Cos[2 δ] Sin[λ] Cos[ϕ])==0,
0<=ϕ<2 π, 0<=δ<=π/2, 0<=λ<2 π
},{ϕ,δ,λ},Method->Reduce]


{{ϕ->0,δ->π/2,λ->0},{ϕ->π/2,δ->π/2,λ->π},{ϕ->π,δ->π/2,λ->0},{ϕ->(3 π)/2,δ->π/2,λ->π}}

• You are very clear.Thank you very much.I was neglected the "Method->Reduce". – yonghui wang May 14 '18 at 6:58