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f(x)=cos(x)⋅cos(2x)⋅cos(3x)⋯cos(999x)\large f(x) = \cos(x) \cdot \cos(2x) \cdot \cos(3x)\cdots \cos(999x)f(x)=cos(x)⋅cos(2x)⋅cos(3x)⋯cos(999x)
If f(2π1999)=12kf \left(\dfrac{2\pi }{1999}\right) = \dfrac{1}{2^{k}}f(19992π)=2k1, find kkk.
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