How do you solve $sin x - 2 sin x cos x = 0$ $sinx-2sinxcosx=0$ ?

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Answer 1 · 2016-09-26T16:36:27

$x = n π$ $x=npi$ or $x = 2 n π \pm \frac{π}{3}$ $x=2npi+-pi/3$ , where $n$ $n$ is an integer

$sin x - 2 sin x cos x = 0$ $sinx-2sinxcosx=0$

$\Leftrightarrow sin x (1 - 2 cos x) = 0$ $hArrsinx(1-2cosx)=0$

Hence either $sin x = 0$ $sinx=0$ i.e. $x = n π$ $x=npi$ , where $n$ $n$ is an integer

or $1 - 2 cos x = 0$ $1-2cosx=0$ i.e. $cos x = \frac{1}{2}$ $cosx=1/2$ and $x = 2 n π \pm \frac{π}{3}$ $x=2npi+-pi/3$ , where $n$ $n$ is an integer

How do you solve sinx-2sinxcosx=0sinx−2sinxcosx=0sinx-2sinxcosx=0?