How can I get analytical solution of trigonometric equations?

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Mukul
Mukul on 20 Jun 2018
Commented: Walter Roberson on 28 Jun 2018
the constants are:
k11 = (16*V1*V1)/(n^3*(pi)^2*(2*pi*f)*L)
k22 = (16*V2*V2)/(n^3*(pi)^2*(2*pi*f)*L)
k33 = (16*V3*V3)/(n^3*(pi)^2*(2*pi*f)*L)
k12 = (8*V1*V2)/(n^3*(pi)^2*(2*pi*f)*L)
k13 = (8*V1*V3)/(n^3*(pi)^2*(2*pi*f)*L)
k23 = (8*V2*V3)/(n^3*(pi)^2*(2*pi*f)*L)
The equations are:
P1 = (k12.*cos(x(1)*pi/360).*cos(x(2)*pi/360).*sin(x(4)*pi/180))+(k13.*cos(x(1)*pi/360).*cos(x(3)*pi/360).*sin(x(5)*pi/180))
P2 = -(k12.*cos(x(1)*pi/360).*cos(x(2)*pi/360).*sin(x(4)*pi/180))+(k23.*cos(x(2)*pi/360).*cos(x(3)*pi/360).*sin((x(5)-x(4))*pi/180))
P3 = -(k13.*cos(x(1)*pi/360).*cos(x(3)*pi/360).*sin(x(5)*pi/180))+(k23.*cos(x(2)*pi/360).*cos(x(3)*pi/360).*sin((x(4)-x(5))*pi/180))
Q1 = (k11.*cos(x(1)*pi/360).*cos(x(1)*pi/360))-(k12.*cos(x(1)*pi/360).*cos(x(2)*pi/360).*cos(x(4)*pi/180))-(k13.*cos(x(1)*pi/360).*cos(x(3)*pi/360).*cos(x(5)*pi/180))
Q2 = -(k12.*cos(x(1)*pi/360).*cos(x(2)*pi/360).*cos(x(4)*pi/180))+(k22.*cos(x(2)*pi/360).*cos(x(2)*pi/360))-(k23.*cos(x(2)*pi/360).*cos(x(3)*pi/360).*cos((x(5)-x(4))*pi/180))
Q3 = -(k13.*cos(x(1)*pi/360).*cos(x(3)*pi/360).*cos(x(5)*pi/180))-(k23.*cos(x(2)*pi/360).*cos(x(3)*pi/360).*cos((x(5)-x(4))*pi/180))+(k33.*cos(x(3)*pi/360).*cos(x(3)*pi/360))
How can I solve for the angles x(1), x(2), x(3), x(4) and x(5)? Can anyone please help me to solve these equations?
  8 Comments
Mukul
Mukul on 26 Jun 2018
Ok Walter, please let me inform what you get when the computation would finish

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Accepted Answer

Walter Roberson
Walter Roberson on 22 Jun 2018
Analytic solution:
x(1) = 180 + 360*Z1
x(2) = 180 + 360*Z2;
x(3) = 180 + 360*Z3;
x(4) and x(5) arbitrary (that is, the above 3 together solve all 5 equations)
Here, Z1, Z2, and Z3 represent arbitrary integers
  6 Comments
Walter Roberson
Walter Roberson on 28 Jun 2018
Solving for x(4) and x(5) both failed at the place I was indicating was taking a long time. I did not go back to try substituting in the other choices.

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