Solving $\left(\;1-a\cos(\theta-\alpha)\;\right)\left(\;1-b\cos(\theta-\beta)\;\right)=\frac14\left(1-a^2\right)\left(1-b^2\right)$ for $\theta$ Let $0\lt a, b\lt 1$ be two constants.
Then, how can I solve the trig equation 
$$\left(\;1-a\cos(\theta-\alpha)\;\right)\left(\;1-b\cos(\theta-\beta)\;\right)=\frac14\left(1-a^2\right)\left(1-b^2\right)$$ for $\theta$ in terms of $a$, $b$ and $\alpha$, $\beta$?
 A: Use angle difference formulas to establish
$$\cos(\theta-\alpha)=\cos\theta\cos\alpha+\sin\theta\sin\alpha$$
Use the tangent half-angle substitution to write
$$\cos\theta=\frac{1-t^2}{1+t^2}\qquad\sin\theta=\frac{2t}{1+t^2}$$
Then you have
$$
\tfrac14\left(1-a^2\right)\left(1-b^2\right)=
\left\{1-a\left(\frac{1-t^2}{1+t^2}\cos\alpha+\frac{2t}{1+t^2}\sin\alpha\right)\right\}\left\{1-b\left(\frac{1-t^2}{1+t^2}\cos\beta+\frac{2t}{1+t^2}\sin\beta\right)\right\}
$$
Multiply by $1+t^2$ to get
$$\tfrac14(1+t^2)\left(1-a^2\right)\left(1-b^2\right)=
\left\{1+t^2-a\left((1-t^2)\cos\alpha+2t\sin\alpha\right)\right\}\left\{1+t^2-b\left((1-t^2)\cos\beta+2t\sin\beta\right)\right\}$$
This is a quartic (i.e. degree $4$) polynomial equation in $t$, which may or may not simplify in some way. Solve it, then reverse the tangent half-angle substitution using $\theta=2\arctan t$. If the degree of your equation should happen to be less than four, be aware of the fact that the tangent half-angle substitution associates $\theta=\pm180°$ with $t=\infty$, so check whether that's a solution as well.
