How do you solve $sinx-2sinxcosx=0$ between the interval $0<=x<=2pi$ ?

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Answer 1 · 2016-11-26T11:53:57

Solution: in interval $0 <= x <= 2pi , x=pi/3, x= (5pi)/3$

$sinx-2sinxcosx =0 or cancelsinx = 2 cancelsinxcosx or 2cosx = 1 or cosx=1/2$
We know $cos (pi/3)=1/2 and cos (2pi-pi/3)=1/2 :. x=pi/3, x= (5pi)/3$

Solution: in interval $0 <= x <= 2pi , x=pi/3, x= (5pi)/3$ [Ans]

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