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March 25, 2017

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Find an equation of the tangent line to the given curve at the given point

y=(1/cosx)-2cosx at((pi/3),1)

y'=0+2sinx
slope = √3

y=mx+b
1=(√3)(pi/3)+b
b=(3-√3pi)/3
3√3x-3y+3-√3pi=0

What did I do wrong? Thanks in advance.

  • Calculus - ,

    Got it.

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