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

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(D) Given equation: $e^{\sin x} - e^{-\sin x} - 4 = 0$Let $y = e^{\sin x}$. Since $e^{-\sin x} = \frac{1}{y}$,the equation becomes:$y - \frac{1}{y} -

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None

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(D) Given equation: $e^{\sin x} - e^{-\sin x} - 4 = 0$Let $y = e^{\sin x}$. Since $e^{-\sin x} = \frac{1}{y}$,the equation becomes:$y - \frac{1}{y} - 4 = 0$Multiplying by $y$ $(y 
eq 0)$:$y^2 - 4y - 1 = 0$Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$y = \frac{4 \pm \sqrt{16 - 4(1)(-1)}}{2} = \frac{4 \pm \sqrt{20}}{2} = 2 \pm \sqrt{5}$Since $y = e^{\sin x} > 0$,we must have $y = 2 + \sqrt{5}$ (as $2 - \sqrt{5} Taking the natural logarithm on both sides:$\sin x = \ln(2 + \sqrt{5})$We know that $\sqrt{4} Since $e^1 \approx 2.718$,$\ln(2 + \sqrt{5}) > \ln(e) = 1$.Since $\sin x \leq 1$ for all real $x$,the equation $\sin x = \ln(2 + \sqrt{5})$ has no real solution.Therefore,the number of real roots is $0$ (None).

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

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