HI Cengizhan,
MODIFIED
The first equation above is
(-1/2)*(cos(x)*cos(z)+sin(theta)/psi) = 0 (1)
which has an extraneous factor of (-1/2) and might as well be written as
(cos(x)*cos(z)+sin(theta)/psi) = 0
How confident are you that this is the correct equation? When I posted this answer intially, without thinking about it I assumed the equation was
(-1/2)*(cos(x)*cos(z)) +sin(theta)/psi = 0 (2)
because this form is the same as that of the remaining two equations. (It did not help that the equations were posted as an image rather than a text file and could not be directly copied).
If you start dividing equations by hand and so forth, for (2) you can arrive at
psi = 6
phi = 1/3
lam = phi*psi;
lam2 = lam^2;
sig2 = (psi^2/4)/(1+lam2);
z = atand(lam)
theta = atand(sqrt((sig2-lam2)/(1-sig2)))
x = atand(-lam/tand(theta))
cosd(z)*cosd(x) - 2*sind(theta)/psi
cosd(x)*sind(z) - 2*phi*sind(theta)
cosd(z)*sind(x) + 2*phi*cosd(theta)
For (1), a solution is
psi = 6
phi = 1/3
lam = phi*psi;
lam2 = lam^2;
sig2 = (psi^2)/(1+4*lam2);
z = atand(-2*lam)
theta = -atand(sqrt((sig2-4*lam2)/(1-sig2)))
x = atand(2*lam/tand(theta))
cosd(z)*cosd(x) + sind(theta)/psi
cosd(x)*sind(z) - 2*phi*sind(theta)
cosd(z)*sind(x) + 2*phi*cosd(theta)
With the use of atand, all angles are -90<angle<90 and the cosines of the angles are all positive.
For the x = pi/2 solution, put cos(x) = 0 into the equations, which forces sin(theta) = 0. Then sin(x) = +-1 and cos(theta) = +-1. So it appears that z = acos(+-2*phi), depending how some of the signs are chosen. Chasing around minus signs for some of the special cases can get tedious.
The algebra above uses atan and I can't really comment on the acos solutions in your comment below. Except that y = acos(1/sqrt(1+tan(y)^2)) where the square root can be of either sign, and algebraic expressions for tangent can be plugged into this.
There are only so many values of phi and psi that result in real angles. For many values of phi and psi the solution is complex. There are regions of the phi-psi plane that don't lead to real angles, and you can't "solve" that issue.
Best Answer