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#.
