A geometry problem by Green Elephant

We divide $\sin^{2}\alpha-2\cos^{2}\alpha=\sin\alpha\cos\alpha$ by $\cos^{2}\alpha$ (which is obviously different that 0 given that $5<\alpha<6$ ) and we obtain: $\large \tan^{2}\alpha-2=\tan\alpha$ . We substitute $\tan\alpha=t$ and we get a second degree equation : $t^{2}-t-2=0$ . The solutions that resut from this equation are $t_{1}=-1$ and $t_{2}=2$ . Given that $5<\alpha<6 \rightarrow \frac{3\pi}{2}<\alpha<2\pi \rightarrow \alpha<0$ Therefore the answer is $\boxed{ -1}$ .