How do you solve: $sin 2 x - sin x = 0$ $sin2x - sinx = 0$ ?

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Answer 1 · 2018-05-28T05:05:40

$x = 0, x = \frac{π}{3}$ $color(maroon)(x = 0 , color(crimson)(x = pi/3$ or $60^{\circ}$ $color(crimson )(60^@$

$sin 2 x - sin x = 0$ $sin 2x - sin x = 0$

$\Rightarrow 2 sin x cos x - sin x = 0$ $=> 2 sin x cos x - sin x = 0$

$\Rightarrow sin x \cdot (2 cos x - 1) = 0$ $=> sin x * (2 cos x - 1) = 0$

$\Rightarrow sin x = 0, cos x = \frac{1}{2}$ $=> sin x = 0, cos x = 1/2$

$x = 0, x = \frac{π}{3}$ $color(maroon)(x = 0 , color(crimson)(x = pi/3$ or $60^{\circ}$ $color(crimson )(60^@$

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 $k$

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