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

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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

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