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[itex]\frac{d\theta1}{dt}[/itex] = [itex]\Omega[/itex] + sin [itex]\theta1[/itex] + [itex]\frac{1}{2}[/itex] (sin[itex]\theta[/itex]1 + sin[itex]\theta[/itex]2)

[itex]\frac{d\theta2}{dt}[/itex] = [itex]\Omega[/itex] + sin [itex]\theta2[/itex] + [itex]\frac{1}{2}[/itex] (sin[itex]\theta[/itex]1 + sin[itex]\theta[/itex]2)

I know that if the absolute value of omega is less than two, there will be 4 critical points, and if the absolute value of omega is greater than two, there will be no critical points. I have found the Jacobian,

[itex]\frac{3}{2}[/itex] cos[itex]\theta1[/itex] [itex]\frac{1}{2}[/itex] cos[itex]\theta2[/itex]

[itex]\frac{1}{2}[/itex] cos[itex]\theta1[/itex] [itex]\frac{3}{2}[/itex] cos[itex]\theta2[/itex]

My problem is that I do not understand how to linearize this system of equations when I do not know the value of omega, because without knowing the value of omega, I cannot solve for [itex]\theta1[/itex] or [itex]\theta2[/itex]. Any advice regarding how to linearize this system, and from linearizing the system classify the fixed points, would be appreciated. Thanks in advance!