I have come across a system of trigonometric equations in $t$, with parameters $x$ and $y$. Though I know the first is not analytically solvable, my hope was to solve the second, and then substitute into the first. However, I am unable to solve the second equation, though I think it may be possible. Without further ado, the equations are: $$\left\{\begin{array}{l}\sin(t)\cos(t)-y\sin(t)-t+x=0\\\cos(2t)-y\cos(t)-1=0\end{array}\right.$$ Indeed, if one considers the first and second left hand sides, one can observe that the second is the derivative of the first with respect to $t$. Not sure if that helps.
Thanks a lot, and any help is appreciated.
Well, since $\cos(2t)=\cos^2t-\sin^2t=2\cos^2 t-1$, we have $$\cos(2t)+y\cos t-1=2\cos^2t-1+y\cos t-1\\ =2\cos^2 t-y\cos t-2=0$$ therefore $$\cos t=\frac{y\pm\sqrt{y^2+16}}{4} $$ or $$t=k\pi\pm\cos^{-1}\frac{y\pm\sqrt{y^2+16}}{4}$$ for $k\in\mathbb{Z}$.