I am asked to solve the following equation: $6\sin(t)=\frac{\cos(2t)-5}{\sqrt{\tan(X)}}$ where X is the solution of $8\sin(2x)+\cos(2x)=10\cot(x)-2$.
I have already tried to replace the cotangent by equivalent expressions but I have not been able to quite solve it the way it is actually done in the textbook. Indeed, the way it is presented is by setting $z=\tan(x)$ and solving $z^3+6z^2+3z-10=0$, thus finding z to be either -5, -2 or 1. I cannot get to this last 3rd-order equation. Any hint would be much appreciated!
Thanks for your help!