Sol. esin x – e –sin x – 4 = 0
Let esin x = y then e-sin x = 1/y
∴ Equation becomes, y – 1/y – 4 = 0
⇒ y2 – 4y – 1 = 0
⇒ y = 2 + √5,2 - √5
But y is real +ve number,
∴ y ≠ 2 - √5 ⇒ y = 2 + √5
⇒ esin x = 2 + √5 ⇒ sin x = loge (2 + √5)
But 2 + √5 > e ⇒ loge (2 + √5) > logee
⇒ loge (2 + √5) > 1 Hence, sin x > 1
Which is not possible.
∴ Given equation has no real solution