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Answer 1 · 2017-12-06T16:19:51

First, we need to know that $sin (2 x) = 2 sin (x) cos (x)$ $sin(2x)=2sin(x)cos(x)$

$sin (2 x) - sin (x) = 0$ $sin(2x)-sin(x)=0$
$2 sin (x) cos (x) - sin (x) = 0$ $2sin(x)cos(x)-sin(x)=0$
$(sin x) (2 cos x - 1) = 0$ $(sinx)(2cosx-1)=0$

we have $sin x = 0 or cos x = \frac{1}{2}$ $sinx=0 or cosx=1/2$

When $sin x = 0$ $sinx=0$ , $x = 0 or x = π$ $x=0 or x=pi$
When $cos x = \frac{1}{2}$ $cosx=1/2$ , $x = \frac{π}{3} or x = \frac{5 π}{3}$ $x=pi/3 or x=(5pi)/3$

Therefore, the answer is not only $0 and π$ $0 and pi$ , but also include $\frac{π}{3} and \frac{5 π}{3}$ $pi/3 and (5pi)/3$ .

Here is the answer :) Hope it can help you.

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