Find the general solution of the given equation sin2a=cos3a

given,

sin2A=cos3A

2 sinA cosA=4cos3A-3cosA

2 sin A=4 cos2A-3

2 sin A =4(1-sin2A)-3

now solve for sinA

sin A=(-1-√5)/4  or (-1+√5)/4

Put A=18

and for 0

sin18=(-1+√5)/4