<p>The equation \(e^{\sin x} - e^{-\sin x} - 4 = 0\) has</p>
<p>infinite number of real roots.</p>
<p>no real roots.</p>
<p>exactly one real root.</p>
<p>exactly four real roots.</p>
Step-by-Step Solution
Key Concept: Recognize that e^(sin x) - e^(-sin x) = 2sinh(sin x), and since sinh is strictly increasing with range ℝ, we need sinh(sin x) = 2. However, since |sin x| ≤ 1, we have |sinh(sin x)| ≤ sinh(1) ≈ 1.175 < 2, making the equation impossible to satisfy.
<p><strong>Step 1:</strong> Rewrite the equation using the identity e^t - e^(-t) = 2sinh(t):</p><p>e^(sin x) - e^(-sin x) - 4 = 0</p><p>⟹ 2sinh(sin x) = 4</p><p>⟹ sinh(sin x) = 2</p><p><strong>Step 2:</strong> Recall that sinh is a strictly increasing function. For sinh(sin x) = 2, we need sin x = sinh⁻¹(2).</p><p><strong>Step 3:</strong> Calculate sinh⁻¹(2) = ln(2 + √5) ≈ 1.444</p><p><strong>Step 4:</strong> Since the range of sine function is [-1, 1], and sinh⁻¹(2) ≈ 1.444 > 1, there is no real value of x satisfying this equation.</p><p><strong>Step 5:</strong> We can verify: maximum value of sinh(sin x) occurs at sin x = 1, giving sinh(1) ≈ 1.175, which is less than 2.</p><p>∴ Answer: <strong>B (No real solution)</strong></p>
Correct Answer: B