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

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(D) Given equation: ${e^{\sin x}} - {e^{-\sin x}} - 4 = 0$.Let ${e^{\sin x}} = y$. Since ${e^{\sin x}} > 0$,we have $y > 0$.The equation becomes $y -

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None

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(D) Given equation: ${e^{\sin x}} - {e^{-\sin x}} - 4 = 0$.Let ${e^{\sin x}} = y$. Since ${e^{\sin x}} > 0$,we have $y > 0$.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 $y = \frac{4 \pm \sqrt{16 - 4(1)(-1)}}{2} = \frac{4 \pm \sqrt{20}}{2} = 2 \pm \sqrt{5}$.Since $y > 0$ and $\sqrt{5} \approx 2.236$,we must have $y = 2 + \sqrt{5}$.Thus,${e^{\sin x}} = 2 + \sqrt{5}$.Taking the natural logarithm on both sides: $\sin x = \ln(2 + \sqrt{5})$.Since $\sqrt{5} > 1$,$2 + \sqrt{5} > 3$. Therefore,$\ln(2 + \sqrt{5}) > \ln(3) > 1$.However,the range of $\sin x$ is $[-1, 1]$.Since $\ln(2 + \sqrt{5}) > 1$,there is no real value of $x$ that satisfies this equation.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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