Given the following two equations, where $c_1, c_2$ are independent constants
- $10(\cos\theta_1) + 10(\cos\theta_2) + 6(\cos\theta_3) = c_1$
- $10(\sin\theta_1) + 10(\sin\theta_2) + 6(\sin\theta_3) + 8= c_2$
Is it even possible to somehow simplify this system of equations to solve for $\theta_1, \theta_2, \theta_3$ ? I have tried using sum to product but it seems like a dead end.
Can anyone confirm that the only way to arrive at solutions for this is to use a optimization program?
Partial answer :Let $(\theta_1 +\theta_2)=m$ and $(\theta_1-\theta_2)/2=d.$ We have $\theta_1=m+d$ and $\theta_2=m-d$ from which $\cos \theta_1 +\cos \theta_2= 2(\cos m )(\cos d)$ and $\sin \theta_1+\sin \theta_2=2 (\sin m)(\cos d)$.So we have $$20(\cos m)\cos d=U$$ $$\text{and }20(\sin m)\cos d=V$$ where $$U=c_1-6\cos \theta_3$$ $$\text{and }V=c_2-8-6\sin \theta_3.$$ Now $U=V=0$ is possible for some $\theta_3$ iff $36=c_1^2+(8-c_2)^2$.In this case we must have $\cos d=0$ because $\cos m, \sin m$ cannot both be $0$.This means that $\theta_1$ and $\theta_2$ can be any values differing by an odd multiple of $\pi$, while $\cos \theta_3=c_1/6$ and $\sin \theta_3=(8-c_2)/6$....In general if $U$ and $V$ are any values for which $$(U-c_1)^2+(V-c_2+8)^2=36$$ then $\theta_3$ can exist. Now for $U,V$ not both $0$ we have $$20\ge|20 \cos d|=\sqrt{U^2+V^2}$$ which appears to place another constraint on $c_1,c_2$, And we have $$ \tan m=V/U \text{ or }\cot m=U/V.$$ Given allowable values for $c_1,c_2$ we can find values for $d,m$ from the above two equations, and obtain $\theta_1=m+d$ and $\theta_2=m-d$.