x=-3.141644, 0.680598 and 3.141541 (-3.142, 0.681 and 3.142 to nearest thousandth).
The negative solution is slightly less than -(pi) but the larger positive solution is less than (pi). So there are two solutions in range.
When x becomes large (positive or negative), the exponential becomes small so that it approaches zero, and sin(x) approaches zero, so x approaches n(pi) where n is an integer, as can be seen by the solutions close to +(pi). For 0<x<1 the exponential is between 1 and 1/e (0.368).
When x is small sin(x) approaches x. e^(-x^2)=1-x^2+x^4/2-x^6/4+...
So approximately x=1-x^2 if subsequent powers can be ignored. x^2+x-1=0 has the positive solution (sqrt(5)-1)/2=0.618.
sin(x)=x-x^3/3!+x^5/5!-... = 1-x^2+x^4-...
Knowing that x is around 0.6, we can find out what power of x is less than 0.0005 (for a solution correct to the nearest thousandth). 0.6^a<0.0005; alog(0.6)=log(0.0005) so a=15. So to solve for x we would need to solve the polynomial as far as x^15.
Another approach is to use the fact that the graphs of sin(x) and e^(-x^2) are fairly straight in the vicinity of x=0.6 and we can consider the intersection of two straight lines. To do this we need to find the gradient at x=0.6. For sin(x) the gradient is cos(x)=cos(0.6)=0.8253. For the exponential, it is -2xe^(-x^2)=-0.8372. y1=0.8253x+c1 and y2=-0.8372x+c2. sin(0.6)=0.5646=y1 and e^(-0.36)=0.6977=y2. c1=0.5646-0.8253*0.6=0.0694; c2=0.6977+0.8372*0.6=1.2000.
y=0.8253x+0.0694=-0.8372x+1.2; 1.6625x=1.2-0.0694=1.1306 so x=0.6801. This corresponds very well to the solution above, which was obtained by trial and error.