Find all solutions if 0<x<2pie : Sin2x(Cosx)+Cos2x(Sinx) = 0

In radians exact values only

sin(2x+x)=0 , 3x=k*pi where k is an integer , x= k*pi/3 , x=pi/3 or 2pi/3 or pi or 4pi/ or 5pi/3

Is this sin(2x)cos(x) + cos(2x)sin(x)
or
sin(x)^2 * cos(x) + cos(x)^2 * sin(x)?


sin(2x)cos(x) + cos(2x)sin(x) = 0
sin(2x + x) = 0
sin(3x) = 0
3x = pi * k
x = (pi/3) * k


sin(x)^2 * cos(x) + cos(x)^2 * sin(x) = 0
sin(x)cos(x) * (sin(x) + cos(x)) = 0

sin(x) = 0
x = pi * k

cos(x) = 0
x = pi/2 + pi * k
x = (pi/2) * (1 + 2k)

sin(x) + cos(x) = 0
sin(x) = -cos(x)
tan(x) = -1
x = 3pi/4 + pi * k
x = (pi/4) * (3 + 4k)


x = pi * k , (pi/2) * (1 + 2k) , (pi/4) * (3 + 4k)

lol its pi not pie