How can I convert x, y, and z which are functions of theta to theta function of x, y, and z?

Hello,

I have three equations,

eqn1 = 2*L*(y+a)*cos(theta1) + 2*z*L*sin(theta1) + x^2 + y^2 + z^2 + a^2 + L^2 + 2*y*a - l^2 == 0

eqn2 = -L*(sqrt(3)*(x+b)+y+c)*cos(theta2) + 2*z*L*sin(theta2) + x^2 + y^2 + z^2 + b^2 + c^2 + L^2 + 2*x*b + 2*y*c - l^2 == 0

eqn3 = L*(sqrt(3)*(x-b)-y-c)*cos(theta3) + 2*z*L*sin(theta3) + x^2 + y^2 + z^2 + b^2 + c^2 + L^2 - 2*x*b + 2*y*c - l^2 == 0

Every value except theta1,theta2, and theta3 are given.

I want to make the three equations as theta functions having x, y, and z variables.

Like a form of theta1 = .... , theta2 = ....., and theta3 = ....

Please let me know what command I can use to make the conversion.

Answer 1

[DGM]

Something like this

syms theta1 theta2 theta3 x y z L l c b a
eqn1 = 2*L*(y+a)*cos(theta1) + 2*z*L*sin(theta1) + x^2 + y^2 + z^2 + a^2 + L^2 + 2*y*a - l^2 == 0
eqn2 = -L*(sqrt(3)*(x+b)+y+c)*cos(theta2) + 2*z*L*sin(theta2) + x^2 + y^2 + z^2 + b^2 + c^2 + L^2 + 2*x*b + 2*y*c - l^2 == 0
eqn3 = L*(sqrt(3)*(x-b)-y-c)*cos(theta3) + 2*z*L*sin(theta3) + x^2 + y^2 + z^2 + b^2 + c^2 + L^2 - 2*x*b + 2*y*c - l^2 == 0

e1 = theta1==solve(eqn1,theta1)
e2 = theta2==solve(eqn2,theta2)
e3 = theta3==solve(eqn3,theta3)

[Hwajin Choi]

That's exactly what I want! I appreciate you.

[DGM]

If it works, just accept the answer so it's taken out of the queue.

[Hwajin Choi]

Opps Sorry I forgot it. Thanks for the reminder!