What is the sollution of sin2x-sin3x?I got two solutions +-Pi/5 and +-3Pi/5 ...but I can't get one more, that is kPi...

We'll re-write what the enunciation provided in a proper
manner:

sin 2x - sin 3x =
0

We'll transform the difference of sines into a product,
to solve the equation:

sin a - sin b =
2cos[(a+b)/2]*sin[(a-b)/2]

sin 2x - sin 3x =
2cos[(2x+3x)/2]*sin[(2x-3x)/2]

sin 2x - sin 3x =
2cos[(5x)/2]*sin[(-x)/2]

We'll re-write the
equation:

2cos[(5x)/2]*sin[(-x)/2] =
0

We'll divide by
2:

cos[(5x)/2]*sin[(-x)/2] =
0

We'll cancel each
factor:

cos[(5x)/2] = 0

5x/2 =
+/-arccos 0 + 2kpi

5x/2 = +/-(pi/2) +
2kpi

5x = +/-(pi) + 4kpi

x =
+/-(pi/5) + 4kpi/5

sin[(-x)/2] =
0

x/2 = (-1)^k*arcsin 0 +
kpi

x/2 = kpi

x =
2kpi

The solutions of the equation are:
{-(pi/5) + 4kpi/5}U{(pi/5) + 4kpi/5}U{2kpi}.