The equation e^{sin x} - e^{-sin x} = 4 has . | vedclass

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(D) Let $e^{\sin x} = y$. Since $e^{\sin x} > 0$,we have $y > 0$. The equation becomes $y - \frac{1}{y} = 4$,which simplifies to $y^2 - 4y - 1 = 0$. S

Vedclass · Accepted Answer

no

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(D) Let $e^{\sin x} = y$. Since $e^{\sin x} > 0$,we have $y > 0$. The equation becomes $y - \frac{1}{y} = 4$,which simplifies to $y^2 - 4y - 1 = 0$. Solving for $y$ using the quadratic formula,$y = \frac{4 \pm \sqrt{16 - 4(1)(-1)}}{2} = \frac{4 \pm \sqrt{20}}{2} = 2 \pm \sqrt{5}$. Since $y > 0$,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 $\sqrt{5} \approx 2.236$,$2 + \sqrt{5} \approx 4.236$. We know that $e \approx 2.718$,so $\ln(4.236) > \ln(e) = 1$. Therefore,$\sin x > 1$,which is impossible for any real $x$. Hence,the equation has no real solutions.

The equation $e^{\sin x} - e^{-\sin x} = 4$ has . . . . . . solutions.

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