How can I get analytical solution of trigonometric equations?

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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?
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
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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