The number of solutions, of the equation $$e^{\sin x}-2 e^{-\sin x}=2$$, is :

For $$\alpha, \beta \in(0, \pi / 2)$$, let $$3 \sin (\alpha+\beta)=2 \sin (\alpha-\beta)$$ and a real number $$k$$ be such that $$\tan \alpha=k \tan \beta$$. Then, the value of $$k$$ is equal to

$$96\cos {\pi \over {33}}\cos {{2\pi } \over {33}}\cos {{4\pi } \over {33}}\cos {{8\pi } \over {33}}\cos {{16\pi } \over {33}}$$ is equal to :