Can anyone please comment on this problem and advise me how to solve it?

# 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?

##### 8 Comments

Walter Roberson
on 26 Jun 2018

There is not necessarily any error in your code. The system is just difficult to solve.

My work so far shows that for each x1 there are two x2, and that for each x2 there are four x3. Computation is slow, so I have not gotten further than that quite yet.

### Accepted Answer

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
on 28 Jun 2018

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