If $$2 \sin ^3 x+\sin 2 x \cos x+4 \sin x-4=0$$ has exactly 3 solutions in the interval $$\left[0, \frac{\mathrm{n} \pi}{2}\right], \mathrm{n} \in \mathrm{N}$$, then the roots of the equation $$x^2+\mathrm{n} x+(\mathrm{n}-3)=0$$ belong to :

The sum of the solutions $$x \in \mathbb{R}$$ of the equation $$\frac{3 \cos 2 x+\cos ^3 2 x}{\cos ^6 x-\sin ^6 x}=x^3-x^2+6$$ is

If $$\alpha,-\frac{\pi}{2}<\alpha<\frac{\pi}{2}$$ is the solution of $$4 \cos \theta+5 \sin \theta=1$$, then the value of $$\tan \alpha$$ is