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

Let $$t=e^{\sin x}$$. Then $$t-2/t=2$$, so $$t^2-2t-2=0$$, $$t=1+\sqrt{3}$$ or $$t=1-\sqrt{3}<0$$ (rejected).

$$e^{\sin x}=1+\sqrt{3}\approx 2.73$$, $$\sin x=\ln(1+\sqrt{3})\approx 1.005>1$$. Since $$|\sin x|\leq 1$$, no solution.

The answer is Option (4): 0.

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