Question

The number of solutions of the equation \(e^{\sin x} - 2e^{-\sin x} = 2\) is

• A. 2
• B. more than 2
• C. 1
• D. 0

Answer 1

The Correct Option is D

Rewrite the equation:

\[ e^{\sin x} - 2e^{-\sin x} = 2. \]

Let \( y = e^{\sin x} \). Then \( e^{-\sin x} = \frac{1}{y} \), and the equation becomes:

\[ y - \frac{2}{y} = 2. \]

Multiply both sides by \( y \) to clear the denominator:

\[ y^2 - 2 = 2y. \]

Rearrange terms:

\[ y^2 - 2y - 2 = 0. \]

This is a quadratic equation in \( y \):

\[ y = \frac{2 \pm \sqrt{4 + 8}}{2} = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}. \]

Since \( y = e^{\sin x} \) and \( e^{\sin x} > 0 \), we discard \( y = 1 - \sqrt{3} \) (as it is negative) and consider \( y = 1 + \sqrt{3} \).

However, for \( y = e^{\sin x} = 1 + \sqrt{3} \), we need \( \sin x = \ln(1 + \sqrt{3}) \). Since \( \ln(1 + \sqrt{3}) \) exceeds the range of \( \sin x \) (which is \([-1, 1]\)), there is no value of \( x \) that satisfies this equation.

Conclusion: There are no solutions.

Thus, the answer is: 0