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(sinx)^2+sin2x=0

We know that sin2x= 2sinx*cosx

==> (sinx)^2+2sinx*cosx=0

==> sinx(sinx+2cosx)=0

Then sinx=0   or  sinx+2cosx=0

First sinx=0

==> x= n*pi , (n=0,1,2,...)

For the other part:

sinx+2xosx=0

==> sinx=-2cosx

==> sinx/cosx= -2

==> tanx=-2

==> x= (n+1)pi-arctan 2

Theb x values are: {n*pi, (n+1)*pi-arctan 2}

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(sinx)^2+sin2x=0

We know that sin2x= 2sinx*cosx

==> (sinx)^2+2sinx*cosx=0

==> sinx(sinx+2cosx)=0

Then sinx=0   or  sinx+2cosx=0

First sinx=0

==> x= n*pi , (n=0,1,2,...)

For the other part:

sinx+2xosx=0

==> sinx=-2cosx

==> sinx/cosx= -2

==> tanx=-2

==> x= (n+1)pi-arctan 2

Theb x values are: {n*pi, (n+1)*pi-arctan 2}

Approved by eNotes Editorial Team