Question

Solve by factoring: 2sinxcosx=sinx in [0,2π)

Answer 1

2sin(x)cos(x) = sin(x)
2sin(x)cos(x) - sin(x) = 0
sin(x)(2cos(x) - 1) = 0
sin(x) = 0
x = 0, pi, 2pi
2cos(x) - 1 = 0
cos(x) = 1/2
x = pi/3,  5π/pi
x = 0, pi/3, pi, 5 pi/3,  2 pi
hope it helps