Key Concepts and Formulas

• Substitution: Simplifying equations by replacing complex expressions with simpler variables.
• Exponential Function Properties: $e^x > 0$ for all real $x$, and $e^{-x} = \frac{1}{e^x}$.
• Range of Sine Function: $-1 \le \sin x \le 1$ for all real $x$.

Step-by-Step Solution

Step 1: Substitute $y = \sin x$ to simplify the equation. We introduce the substitution $y = \sin x$ to simplify the given equation. Since $-1 \le \sin x \le 1$, we have $-1 \le y \le 1$.

The equation becomes: $e^y - e^{-y} - 4 = 0$

Step 2: Substitute $t = e^y$ to further simplify the equation. Let $t = e^y$. Since $e^y > 0$ for all real $y$, we have $t > 0$. Also, since $-1 \le y \le 1$, we have $e^{-1} \le t \le e^1$. We also know that $e^{-y} = \frac{1}{e^y} = \frac{1}{t}$.

The equation now becomes: $t - \frac{1}{t} - 4 = 0$

Step 3: Solve the quadratic equation for $t$. Multiply the equation by $t$ to eliminate the fraction: $t^2 - 1 - 4t = 0$ Rearrange the terms to get a standard quadratic equation: $t^2 - 4t - 1 = 0$ Use the quadratic formula to solve for $t$: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = -4$, and $c = -1$. $t = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(-1)}}{2(1)}$ $t = \frac{4 \pm \sqrt{16 + 4}}{2}$ $t = \frac{4 \pm \sqrt{20}}{2}$ $t = \frac{4 \pm 2\sqrt{5}}{2}$ $t = 2 \pm \sqrt{5}$ So, we have two possible values for $t$: $t_1 = 2 + \sqrt{5}$ $t_2 = 2 - \sqrt{5}$

Step 4: Substitute back and analyze the solutions for $x$.

Case 1: $t = 2 - \sqrt{5}$ $e^{\sin x} = 2 - \sqrt{5}$ Since $\sqrt{5} \approx 2.236$, we have $2 - \sqrt{5} \approx -0.236$. Thus, $2 - \sqrt{5} < 0$. But $e^{\sin x}$ must be positive for all real $x$. Therefore, there are no solutions in this case.

Case 2: $t = 2 + \sqrt{5}$ $e^{\sin x} = 2 + \sqrt{5}$ Taking the natural logarithm of both sides: $\sin x = \ln(2 + \sqrt{5})$ Since $\sqrt{5} \approx 2.236$, we have $2 + \sqrt{5} \approx 4.236$. $\sin x = \ln(4.236)$ We know that $\ln(4.236) \approx 1.44$. $\sin x \approx 1.44$ However, the range of the sine function is $-1 \le \sin x \le 1$. Since $1.44 > 1$, there are no solutions in this case.

Step 5: Determine the number of real roots. Since neither case yields a valid solution for $x$, the original equation has no real roots.

Common Mistakes & Tips

• Always check the range of $\sin x$ ($-1 \le \sin x \le 1$) and $e^z > 0$.
• Be careful with substitutions and remember to substitute back to the original variable.
• Don't forget to consider the domain of the exponential and logarithmic functions.

Summary

By using the substitutions $y = \sin x$ and $t = e^y$, the given equation was transformed into a quadratic equation. Solving for $t$ yielded two possible values, $2 \pm \sqrt{5}$. However, neither of these values leads to a valid solution for $x$ because either $e^{\sin x}$ would be negative or $\sin x$ would be outside the range $[-1, 1]$. Therefore, the original equation has no real roots.

Final Answer The final answer is \boxed{no real roots}, which corresponds to option (B).