Sin2x=2sinx

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

sin2x = 2sinx

==> sin2x-2sinx = 0

sin2x= sin(x+x) = 2sinx cos x

==> 2sinx cos x -2sinx=0

Factorize 2sinx

==> 2sinx(cosx-1)=0

Then the sloutions are:

sinx=0    or  cosx=1

x1= arcsin 0 = n*pi

x2= arccos 1 = npi 

x= {npi : n=0,1,2 ...}

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

First, we'll substitute sin 2x by the equivalent expression 2sin x*cosx 

2 sin x*cos x= 2sin x

We'll move all terms to one side:

2 sin x*cos x - 2 sin x=0

2 sin x (cos x -1)=0

We'll put each factor from the product above as being 0.

2 sin x =0

We'll divide by 2:

sin x=0

This is an elementary equation:

x = arcsin 0

x=0 or x=pi

cos x - 1=0

This is an elementary equation, also:

cos x=1

x=+/-arccos 1

x=0, x=pi

The solutions of the equation are {0,pi}.

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

To find solutions of sin2x = 2sinx

Solution:

Sin2x = 2sinxcosx is a trigonometric identy for all x. Therefore the given identity becomes:

2sinxcosx = 2sinx. Or

2sinxcosx -2sinx = 0.

2sinx(cosx-1) = 0.

So sinx = 0 gives x = npi.

cosx - 1 = 0 gives cosx = 1. Or x = 2npi. Or

x = npi, n = 0,1,2...

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