The equation esin x – e–sin x – 4 = 0 has(a) no real roots

1.	The equation esin x – e–sin x – 4 = 0 has(a) no real roots (b) exactly one real root (c) exactly four real roots (d) infinite number of real roots.
Answer» (a) no real roots Given, e^{sin x} – e^{–sin x} – 4 = 0 Let e^{sin x} = y. Then, the given equation becomes y – \(\frac{1}{4}\) = 4 ⇒ y² – 4 –1 = 0 ⇒ y = \(\frac{4±\sqrt{16+4}}{2}\) = 2 ± √5 ⇒ e^{sin x}= 2 ± √5 ⇒ sin x = log_e(2 ± √5) ⇒ sin x = log_e (2 + √5) (∴ (2 - √5) 0 < and so log_e (2 + √5) − is not defined) Now (2 + √5) > 4 ⇒ log_e (2 - √5) > 1 But the value of sin x lies between –1 and 1, both values inclusive, so sin \(x\) ≠ loge (2 + √5) ∴ There are no possible real roots of the given equation.

The equation esin x – e–sin x – 4 = 0 has(a) no real roots (b) exactly one real root (c) exactly four real roots (d) infinite number of real roots.

Answer»

(a) no real roots

Given, e^{sin x} – e^{–sin x} – 4 = 0

Let e^{sin x} = y. Then, the given equation becomes

y – \(\frac{1}{4}\) = 4 ⇒ y² – 4 –1 = 0

⇒ y = \(\frac{4±\sqrt{16+4}}{2}\) = 2 ± √5

⇒ e^{sin x}= 2 ± √5 ⇒ sin x = log_e(2 ± √5)

⇒ sin x = log_e (2 + √5)

(∴ (2 - √5) 0 < and so log_e (2 + √5) − is not defined)

Now (2 + √5) > 4 ⇒ log_e (2 - √5) > 1

But the value of sin x lies between –1 and 1, both values inclusive, so sin \(x\) ≠ loge (2 + √5)

∴ There are no possible real roots of the given equation.