Given equation is esin‌x−e−sin‌x−4=0 Put esin‌x=t in the given equation, we get t2−4t−1=0 ⇒‌‌t=‌
4±√16+4
2
=‌
4±√20
2
‌‌=‌
4±2√5
2
=2±√5 ⇒‌‌esin‌x=2±√5(∵t=esin‌x) ⇒esin‌x=2−√5‌ and ‌esin‌x=2+√5 ⇒‌‌esin‌x=2−√5<0 ‌ and ‌sin‌x=ln(2+√5)>1 So, rejected. Hence, given equation has no solution. ∴‌‌ The equation has no real roots.