The number of solutions of the equation e^{si | vedclass

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(D) Let $e^{\sin x} = t$. Since $\sin x \in [-1, 1]$,$t$ must be in the range $[e^{-1}, e^1]$,i.e.,$[1/e, e] \approx [0.368, 2.718]$.Given equation: $

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$0$

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(D) Let $e^{\sin x} = t$. Since $\sin x \in [-1, 1]$,$t$ must be in the range $[e^{-1}, e^1]$,i.e.,$[1/e, e] \approx [0.368, 2.718]$.Given equation: $t - \frac{2}{t} = 2$.Multiplying by $t$: $t^2 - 2t - 2 = 0$.Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$t = \frac{2 \pm \sqrt{4 - 4(1)(-2)}}{2} = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}$.Since $t > 0$,we take $t = 1 + \sqrt{3} \approx 1 + 1.732 = 2.732$.We require $e^{\sin x} = 1 + \sqrt{3}$.Since $1 + \sqrt{3} \approx 2.732$ and $e \approx 2.718$,we have $1 + \sqrt{3} > e$.Because the maximum value of $e^{\sin x}$ is $e^1 = e$,the equation $e^{\sin x} = 1 + \sqrt{3}$ has no real solution for $x$.Thus,the number of solutions is $0$.

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

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