trig
posted by arka bhaskar .
sin2x=cos3x then sinx=?(if 0<x<pi/2)

sin 2x = cos3x = sin(pi/2  3x)
Therefore, one solution is
2x = pi/2  3x
5x = pi/2
x = pi/10 = 18 degrees
There is a second solution at x = pi/2, but that is just outside the domain of interest. 
Do you mean
sin^2 x = cos^3 x
or do you mean
sin(2x) = cos(3x)
?????
If you mean sin(2x) = cos(3x)
then
2 sin x cos x = 4 cos^3 x  3 cos x
2 sin x = 4 cos^2 x  3
2 sin x = 4 (1  sin^2 x)  3
2 sin x = 4  4 sin^2 x  3
4 sin^2 x + 2 sin x 1 = 0
let z = sin x
4 z^2 + 2 z  1 = 0
z = [ 2 +/ sqrt (4+16) ] / 8
z = [ 2 +/ 2 sqrt 5 ] /8
or
sin x = (1 +/ sqrt 5)/4
if x in quadrant 1 then use + root
sin x = (sqrt 5  1)/4 
Do it the way Dr WLS did it, much simpler and more likely correct.