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

Answer: (D)

0

Answer 2

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