3 years ago

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

1 answer:

4 0

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

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