sin’e + - 1 + tane

The number of real solutions of sin(e^x) = e^1 + e^(-1) is

Number of solutions to the equation |2e^(sin x) - 3 - e^(2 sin x)|= | e^(sin x) - e^(2 sinx) - 1| i n [0, 2 pi] is

e^(sin h^(-1)(tan theta)) =

e^(sin h^(-1) (cot theta))

From the two e.m.f. Equation e_(1)=E_(0) sin (100 pi t) and e_(2)=E_(0)sin (100 pi t+(pi)/3) , we find that

f1+sin^(^^)@0=3sin e cose , then prove that tane =1 or

(e^(y)+1)cos x dx+e^(y)sin x dy =0 A) (e^(y)+1)sin x =c B) (e^(y)+1)=c tan x C) e^(y)sin x + cos x=c D) (e^(x)+1)sin y=c

int (sin x * e ^ (cos x) - (sin x + cos x) e ^ (sin x + cos x)) / (e ^ (2sin x) -2e ^ (sin x) +1) dx

If e^(sin x)-e^(-sin x)-4=0 , then x= (a) 0 (b) sin^(-1){(log)_e(2+sqrt(5))} (c) 1 (d) none of these