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Find out the solutions of the equation sin 2x = 2 sin x, in the interval (0,2pi).

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greenturtle | Student, Grade 10 | (Level 1) Honors

Posted April 17, 2009 at 1:36 AM via web

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Find out the solutions of the equation sin 2x = 2 sin x, in the interval (0,2pi).

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted May 29, 2009 at 1:40 AM (Answer #1)

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sin 2x= 2 sin x

2 sin x cos x= 2sin x

2 sin x cos x - 2 sin x=0

2 sin x (cos x -1)=0

It's a two factors multiplication, and the result is 0 when one of these two factors is 0, too.

2 sin x =0, sin x=0

x=arcsin 0

x=0 or x=pi

cos x - 1=0

cos x=1

x=+/-arccos 1

x=0, x=pi

 

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jeyaram | Student, Undergraduate | (Level 1) Valedictorian

Posted September 14, 2009 at 6:37 PM (Answer #2)

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sin 2x = 2 sin x 2sinXcosX=2sinX sinX(cosX-1)=0 so sinX=0 or cosX-1=0 X=0 or pi

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