The number of real roots of the equation e^{s | vedclass

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Question

(A) Let $e^{\sin x} = y$. Since $\sin x \in [-1, 1]$,we have $y \in [e^{-1}, e^1]$,which means $y \in [1/e, e]$.The equation becomes $y - \frac{1}{y}

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$0$

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(A) Let $e^{\sin x} = y$. Since $\sin x \in [-1, 1]$,we have $y \in [e^{-1}, e^1]$,which means $y \in [1/e, e]$.The equation becomes $y - \frac{1}{y} - 4 = 0$.Multiplying by $y$,we get $y^2 - 4y - 1 = 0$.Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $y = \frac{4 \pm \sqrt{16 + 4}}{2} = 2 \pm \sqrt{5}$.Since $y > 0$,we consider $y = 2 + \sqrt{5} \approx 2 + 2.236 = 4.236$.However,the maximum value of $y = e^{\sin x}$ is $e^1 \approx 2.718$.Since $4.236 > 2.718$,there is no real value of $x$ such that $e^{\sin x} = 2 + \sqrt{5}$.Thus,the number of real roots is $0$.

The number of real roots of the equation $e^{\sin x} - e^{-\sin x} - 4 = 0$ is:

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