WEBVTT
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Hey, it's clear sailing you right here. So
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we have F of X is equal to X plus
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to sign, so my first derivatives get one plus
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to co sign. We know that has a horizontal
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tangent when the first derivative it's equal to zero.
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So it's worn plus to co sign is equal to
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zero. So we get X is equal to two
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pi thirds or four pi thirds. We put it
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in the form Texas difficult to two pi over three
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was two n pi since co sign has a period
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of two pi Or it could be xs equal to
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four pipe thirds plus two and pie and is any
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integer. You realize that these solutions to pie throwers
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and four pi thirds rpai plus or minus pi over
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three. So to be written, there's two and
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prying was pie, plus or minus pi thirds just
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equal to two end plus one times pi plus or
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minus 5/3 And is any integer