I would like to find an analytical expression of the following transcendental equation (see attached):
a1*sin(theta + phi1)+ a2*sin(2*theta + phi2) = 0
a1 a1 phi1 phi2 → Real
For which value of theta, the equation is equal to zero? (only 1 period, i.e. 0 < theta < 2pi)
Any clue?
Thanks !
Dear Julian, yes, Mathematica gives analitycal expressions for theta, but they have complicated shape; in the case a1 = - a2, the solution is theta = phi1-phi2.
Gianluca
Simplify it to sin(alpha)*sin(Theta)*cos(Phi) + sin(alpha)*cos(Theta)*sin(Phi) + 2*cos(alpha)*sin(Theta)*cos(Theta) = 0,
where theta= -phi2/2+Theta; phi1= Phi+phi2/2; tan(alpha) = a1/a2.
Set Theta = arcsin(X). Simplify, and you'll get a standard 4th order polynomial equation with 4 analytic solutions for X (not really beautiful but bearable). But better use appropriate parametric solution, if your problem allows that.
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