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

Answer 2

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