How do you solve: #sin2x - sinx = 0#?

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Answer 1 · 2018-05-28T05:05:40

#color(maroon)(x = 0 , color(crimson)(x = pi/3# or #color(crimson )(60^@#

#sin 2x - sin x = 0#

#=> 2 sin x cos x - sin x = 0#

#=> sin x * (2 cos x - 1) = 0#

#=> sin x = 0, cos x = 1/2#

#color(maroon)(x = 0 , color(crimson)(x = pi/3# or #color(crimson )(60^@#

Answer 2 · 2018-05-28T13:08:46

# x = 180^circ k or x= \pm 60^circ + 360^circ k quad # integer #k#

#sin 2x - sin x = 0#

#2 sin x cos x - sin x = 0#

#sin x (2 cos x -1) = 0#

#sin x = 0 or cos x = 1/2 #

# x = 180^circ k or x= \pm 60^circ + 360^circ k quad # integer #k#