Here, sin2x-cos3x=0
rarr sin2x=cos3x
rarr 2sinx*cosx=4cos^3x-3cosx
rarr 2sinx*cosx=cosx(4cos^2x-3)
rarr 2sinx=4cos^2x-3
rarr 2sinx=4(1-sin^2x)-3
rarr 2sinx=4-4sin^2x-3
rarr 4sin^2x+2sinx-1=0
Comparing it with ax^2+bx+c=0 we have
a=4 b=2 and c=-1
Using quadratic formula
sinx=(-b+-sqrt(b^2-4ac))/(2*a)
=(-2+-sqrt(2^2-4*4*(-1)))/(2*4)
=(-2+-sqrt(4+16))/8
=(-2+-sqrt(5*2^2))/8
=(-2+-2sqrt(5))/8
Taking + sign we get
sinx=(-2+2sqrt(5))/8
sinx=(sqrt(5)-1)/4
sinx=sin(pi/10)
x=npi+(-1)^npi/10
Taking - sign we get
sinx=(-2-2sqrt(5))/8
sinx=-(sqrt(5)+1)/4
sinx=sin((13pi)/10)
x=npi+(-1)^n(13pi)/10
So the required solution is npi+(-1)^npi/10 and npi+(-1)^n(13pi)/10.