The number of real roots of the equation esin⁡x−e−sin⁡x−4=0 are

. Given equation is esin⁡x−e−sin⁡x−4=0Let esin⁡x=y,then given equation can be written asy2−4y−1=0⇒y=2±5But the value ofy=esin⁡x' is always positive, so we take only y=2+5⇒loge⁡y=loge⁡(2+5)⇒sin⁡x=loge⁡(2+5)>1Which is impossible since sin x  cannot be greaterthan 1.Hence, we cannot find any real value of x which satisfiesthe given equation.