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

To solve the given equation \(e^{\sin x} - 2e^{-\sin x} = 2\), let's first introduce a substitution. Let \(y = e^{\sin x}\). Then \(e^{-\sin x} = \frac{1}{y}\). Substituting these into the equation gives:

\(y - 2\left(\frac{1}{y}\right) = 2\)

Multiplying through by \(y\) to clear the fraction, we get:

\(y^2 - 2 = 2y\)

Rearrange the equation:

\(y^2 - 2y - 2 = 0\)

Now, we solve this quadratic equation using the quadratic formula, \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -2\), and \(c = -2\).

\(y = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1}\)

\(y = \frac{2 \pm \sqrt{4 + 8}}{2}\)

\(y = \frac{2 \pm \sqrt{12}}{2}\)

\(y = \frac{2 \pm 2\sqrt{3}}{2}\)

\(y = 1 \pm \sqrt{3}\)

Thus, the potential solutions for \(y\) are \(1 + \sqrt{3}\) and \(1 - \sqrt{3}\). However, since \(y = e^{\sin x}\), which is always a positive number (as the exponential function is always positive), \(1 - \sqrt{3}\) is not possible because it is negative.

Let's analyze the feasible solution:

For \(y = 1 + \sqrt{3}\), we have:

\(e^{\sin x} = 1 + \sqrt{3}\)

This implies \(\sin x = \ln(1 + \sqrt{3})\). However, the range of the sine function is \([-1, 1]\), and \(\ln(1 + \sqrt{3})\) is greater than 1. Therefore, it is not possible for \(\sin x\) to equal \(\ln(1 + \sqrt{3})\).

As there is no valid solution that satisfies the original equation, the number of solutions is 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

Answer 3

Given: 

Take \( e^{\sin x} = t \) where \( t > 0 \).

Step 1: Solve for \( t \):

We start by solving the equation: \[ t - \frac{2}{t} = 2 \] Multiplying both sides by \( t \): \[ t^2 - 2 = 2t \] Rearranging the terms: \[ t^2 - 2t - 2 = 0 \] \[ t^2 - 2t + 1 = 3 \] Factoring the quadratic: \[ (t - 1)^2 = 3 \] Taking the square root of both sides: \[ t - 1 = \pm \sqrt{3} \] Solving for \( t \): \[ t = 1 \pm 1.73 \] This gives two possible values for \( t \): \[ t = 2.73 \quad \text{or} \quad t = -0.73 \] Since \( t > 0 \), we reject \( t = -0.73 \).

Step 2: Solve for \( \sin x \):

We now have: \[ e^{\sin x} = 2.73 \] Taking the natural logarithm of both sides: \[ \ln e^{\sin x} = \ln 2.73 \] Simplifying: \[ \sin x = \log_e 2.73 \] Using a calculator: \[ \sin x = 1.0 \]

Conclusion:

Since \( \sin x = 1 \), this means that there is no solution for \( x \) because the value of \( \sin x \) cannot exceed 1. Therefore, there is no solution.