# solve sin2x=cos2x+1

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sin2x=2sinxcosx

`cos2x=2cos^2x-1`

`2sinxcosx=2cos^2x-1+1`

`2sinxcosx=2cos^2x`

`2cos^2x-2sinxcosx=0`

`2cosx(cosx-sinx)=0`

So solutions are cosx=0 which is `x=pi/2+npi` or

cosx-sinx=0 or cosx=sinx which is `pi/4+npi`

So the solutions are `x=pi/2+npi` or `x=pi/4+npi`