If e^{sin x}-e^{-sin x}-4=0, then the number | vedclass

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(A) Given that, $e^{\sin x}-e^{-\sin x}-4=0$. Let $e^{\sin x}=t$. Since $e^{\sin x} > 0$, we have $t > 0$. The equation becomes $t - \frac{1}{t} - 4 =

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$0$

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(A) Given that, $e^{\sin x}-e^{-\sin x}-4=0$. Let $e^{\sin x}=t$. Since $e^{\sin x} > 0$, we have $t > 0$. The equation becomes $t - \frac{1}{t} - 4 = 0$, which simplifies to $t^2 - 4t - 1 = 0$. Solving for $t$ using the quadratic formula: $t = \frac{4 \pm \sqrt{16 - 4(1)(-1)}}{2} = \frac{4 \pm \sqrt{20}}{2} = 2 \pm \sqrt{5}$. Since $t > 0$, we must have $t = 2 + \sqrt{5}$ (as $2 - \sqrt{5} Thus, $e^{\sin x} = 2 + \sqrt{5}$. We know that $-1 \leq \sin x \leq 1$, which implies $e^{-1} \leq e^{\sin x} \leq e^1$. Numerically, $e \approx 2.718$ and $e^{-1} \approx 0.368$. Also, $2 + \sqrt{5} \approx 2 + 2.236 = 4.236$. Since $4.236 > 2.718$, the equation $e^{\sin x} = 2 + \sqrt{5}$ has no solution for $x$. Therefore, the number of real values of $x$ is $0$.

If $e^{\sin x}-e^{-\sin x}-4=0,$ then the number of real values of $x$ is

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