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I have equation that I try to solve for one of the values

$\sum_{k=0}^{n-1}\cos(2 \pi fk)(x_{k}- \mu-A\cos(2 \pi fk)-B\sin(2 \pi fk))$ I know to set equal $0$ I try to solve for A but how to take sum ?

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    @nname: By removing the equations from your question, you made it impossible to understand. That constitutes vandalism. I have undone your edits, and if you do something like that again, you may be suspended.2012-06-18

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I think OP wants to set the sum to zero and solve for $A$. So, $\sum_{k=0}^{n-1}\cos(2\pi fk)(x_k-\mu-A\cos(2\pi fk)-B\sin(2\pi fk))=0$ becomes $\sum_{k=0}^{n-1}\cos(2\pi fk)(x_k-\mu-B\sin(2\pi fk))-A\sum_{k=0}^{n-1}\cos^2(2\pi fk)=0$ which gives us $A={\sum_{k=0}^{n-1}\cos(2\pi fk)(x_k-\mu-B\sin(2\pi fk))\over\sum_{k=0}^{n-1}\cos^2(2\pi fk)}$